# Stoichiometry core (SL & HL) paper 1 questions

Topic 1 (1.1, 1.2 and 1.3)

Paper 1 style questions are multiple choice. You are not permitted to use a calculator or the data book for these questions, but you should use a periodic table.

A periodic table pop-up is available on the left hand menu.

1

Complete this question without a calculator.

0.20 mol of hydrochloric acid is mixed with 0.20 mol of calcium carbonate:

2HCl + CaCO3 → CaCl2 + H2O + CO2

Which is correct?

Quantities of both reagents are given, which suggests that one reagent will be in excess and the other will be limiting. This is best considered in terms of the mole ratio:

 2HCl + CaCO3 → CaCl2 + H2O + CO2 2 : 1 : 1 : 1 : 1

2 moles of HCl will react with 1 mole of CaCO3.

Therefore 0.20 moles of HCl will react with 0.10 moles of CaCO3. 0.20 moles of CaCO3 has been used, so the calcium carbonate is in excess, and the hydrochloric acid is therefore the limiting reagent.

2 moles of HCl will yield 1 mole of CO2.

Therefore 0.20 moles of HCl will yield 0.10 moles of CO2.

Thus Limiting reagent is HCl and maximum yield of CO2 is 0.10 mol is the correct answer.

2

What is the sum of all coefficients when the following equation is balanced using the smallest possible whole numbers?

___GeO2 + ___HCl → ___GeCl4 + ___H2O

GeO2 + 4HCl → GeCl4 + 2H2O is the balanced equation, so the total sum is 8 - don't forget that GeO2 and GeCl4 both have a '1' in front of them, although by convention we don't actually write that into the equation.

3

Complete this question without a calculator.

Which contains the greatest number of moles of oxygen atoms?

This question requires us first to work out the number of oxygen atoms in each formula, and to then multiply that by the number of moles, which will give us the number of moles of oxygen atoms:

0.05 mol Ca(NO3)2 : calcium nitrate has 6 oxygen atoms: 6 × 0.05 = 0.30

0.10 mol C6H4(NO2)2 : dinitrobenzene has 4 oxygen atoms: 4 × 0.10 = 0.40

0.20 mol CO : carbon monoxide has 1 oxygen atom: 1 × 0.20 = 0.20

0.10 mol O2 : oxygen has 2 oxygen atom: 2 × 0.10 = 0.20

Thus 0.10 mol C6H4(NO2)2 is the correct answer as this contains the largest number of moles of oxygen atoms (0.40 mol).

4

Which of these is an example of a heterogeneous mixture?

A heterogeneous mixture is a mixture of more than one phase (or state) of matter. Hence Solid calcium carbonate powder suspended in water is the correct answer as this contains both a solid (calcium carbonate powder) and a liquid (water). The other answers are homogeneous (one phase of matter) mixtures (a solution like aqueous sodium chloride is uniform throughout and considered to be homogeneous), or in the case of nitrogen dioxide gas - a pure compound.

5

Complete this question without a calculator.

What is the empirical formula of a hydrocarbon with 90% carbon and 10% hydrogen by mass?

We need to find the simplest ratio of the number of (moles of) atoms. We cannot use a calculator so we will need to approximate the relative atomic masses that we take from the periodic table (e.g. carbon 12.01 = 12).

Start by assuming 100g: Thus 90g of C atoms, 10g of H atoms:

 Carbon Hydrogen moles = mass/molar mass 90/12 = 7.5 10/1 = 10 multiply up to whole numbers (×2) 15 20 find the largest common multiple to get the simplest ratio (5 in this case) 15 ÷ 5 = 3 20 ÷ 5 = 4

Therefore the empirical formula and the correct answer is C3H4

6

Which relationship would give the following graph when plotted for a fixed mass of an ideal gas with all other variables constant?

The ideal gas equation is PV=nRT. The letters represent Pressure, Volume, number of moles, gas constant (R) and Temperature in that order.

P against T and V against T will both give a directly proportional linear relationship as the two variables are on opposite sides of the equals sign in the equation; so any change in one variable must be proportionally mirrored in the other variable if all the other values are constant.

P against $${1\over V}$$ will also give a straight line. If all other values in PV=nRT are constant then PV=constant. If we give that constant a value of 1 (PV=1) then P =$${1 \over V}$$ so volume is inversely proportional to pressure and P against $${1\over V}$$ will give a straight line when plotted.

P against V will give the curve (a hyperbola) as shown in the diagram since (in the same way as above) PV=constant, as pressure is increased volume decreases and vica versa. This inversely proportional relationship will (mathematically) always give a curve when plotted.

(If the maths is too much - learn the laws (Avogadro's, Boyle's, Charles' and Dalton's) on the revision slide.)

Thus P against V is the correct answer.

7

Complete this question without a calculator.

How many moles of Iron will be produced if this reaction produces 500 mol of carbon dioxide?

Fe2O3 + 3CO → 2Fe + 3CO2

The mole ratio must be used to answer this question:

 Fe2O3 + 3CO → 2Fe + 3CO2 1 : 3 : 2 : 3

The reaction gives 2 moles of iron (Fe) for every 3 moles of carbon dioxide (CO2) produced.

Therefore for every 1 mole of CO2 produced, $${2 \over 3}$$ of a mole of Fe is produced.

If 500 moles of CO2 are produced, $${2 \over 3}$$ × 500 = 333 moles of Fe will be produced.

Thus 333 mol is the correct answer.

8

What is the sum of the smallest integer coefficients when ethene undergoes complete combustion?

___C2H4 + ___O2 → ___CO2 + ___H2O

C2H4 + 3O2 → 2CO2 + 2H2O is the balanced equation, so the sum of the integer coefficients is 8 - don't forget that C2H4 does have a '1' in front of it, although by convention we don't actually write that into the equation.

(When balancing combustion reactions of hydrocarbons, first balance out carbon and hydrogen atoms by adding coefficients (numbers) in front of CO2 and H2O, after that balance the oxygen atoms and the hydrocarbon as required.)

9

Complete this question without a calculator.

What is the volume of an ideal gas when the pressure on 100cm3 of gas is changed from 200kPa to 100kPa at constant temperature?

The ideal gas equation is PV=nRT. The letters represent Pressure, Volume, number of moles, gas constant (R) and Temperature in that order.

If all other values in PV=nRT are constant then PV=constant. If we give that constant a value of 1 (PV=1) then P =$${1 \over V}$$ so volume is inversely proportional to pressure (and P against $${1\over V}$$ will give a straight line when plotted).

Therefore, if we halve the pressure, from 200 to 100 kPa, we must double the volume: Thus 200 cm3 is the correct answer.

10

Complete this question without a calculator.

What is the concentration, in mol dm−3, of 10.0g of NaOH (Mr = 40.0) in 500cm3 of solution ?

Concentration (mol dm−3) = moles (mol) / volume of solution (dm3)

Moles (mol) = Mass (g) / molar mass (g mol−1)

So moles of sodium hydroxide (NaOH) = 10.0/40.0 = 0.250

Concentration of the solution = 0.250 / 0.500 (The volume must be converted to dm3: 1 dm3 = 1000cm3)

Concentration = 0.250 / 0.500 = 0.500 mol dm−3

11

Complete this question without a calculator.

What volume of carbon dioxide can be obtained by reacting 500 cm3 of methane (CH4) with 500 cm3 of oxygen (O2)?

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

Ideal gases have a constant molar volume (at any given temperature and pressure).

Quantities of both reagents are given, which suggests that one reagent will be in excess and the other will be limiting. This is best considered in terms of the mole ratio:

 CH4 + 2O2 → CO2 + 2H2O 1 : 2 : 1 : 2

1 mole of CH4 will react with 2 moles of O2.

So methane will react with twice its volume of oxygen (and vice versa, oxygen will react with half its volume of methane).

Methane gas is therefore in excess and oxygen is the limiting reagent.

The amount of methane that will react is therefore 500 cm3 / 2 = 250cm3 (1:2 ratio)

And 250 cm3 of methane reacting with give 250 cm3 of carbon dioxide (CO2) (1:1 ratio)

12

Complete this question without a calculator.

How many moles of CaSO4 are required to produce 128 g of SO2? (Ar: S = 32, O = 16)

3CaSO4 + CaS → 4CaO + 4SO2

The mole ratio must be used to answer this question:

 3CaSO4 + CaS → 4CaO + 4SO2 3 : 1 : 4 : 4

The reaction produces 4 moles of SO2 for every 3 moles of CaSO4 reacted.

Therefore for every 1 mole of SO2 produced, $${3 \over 4}$$ of a mole of CaSO4 must be reacted.

Using Moles (mol) = mass (g) / molar mass (g mol−1) (molar mass of SO2 = 32 + 16×2 = 64) 128g of SO2 is 128 / 64 = 2.0 moles

If 2.0 moles of SO2 is produced, $${3 \over 4}$$ × 2.0 = 1.5 moles of CaSO4 will be required.

Thus 1.5 mol is the correct answer.

13

Complete this question without a calculator.

24 g of bromine react with 18.1 g of metal, M, to form MBr3. What is the relative atomic mass of metal M? (Ar: Br = 80)

The formula of the metal bromide is MBr3. This tells us that one mole of metal atoms will combine with three moles of bromine atoms (1:3).

Using Moles (mol) = mass (g) / molar mass (g mol−1), 24 g of bromine is 24 / 80 = 0.3 moles (the common multiple for mental arithmetic here is 8)

Using the ratio 3:1, 0.3 moles of bromine combines with 0.1 moles of metal, M.

So if 0.1 moles of metal M has a mass of 18.1 g (given in the question) then 1 mole of metal M will have a mass of 18.1 × 10 = 181 g

Therefore the relative atomic mass of metal M is 181.

14

Complete this question without a calculator.

What is the molecular formula of a hydrocarbon containing 85.6% carbon by mass with an integer molar mass of 168g mol−1?

We need to find the simplest ratio of the number of (moles of) atoms. We cannot use a calculator so we will need to approximate the relative atomic masses that we take from the periodic table (e.g. carbon 12.01 = 12).

Start by assuming 100g: Thus 85.6g of C atoms, 14.4g of H atoms:

 Carbon Hydrogen moles = mass/molar mass 85.6/12 = A little over 7 (7×12=84) 14.4/1 = 14.4 Approximate mole ratio 7... 14... Simplest mole ratio (empirical formula) 1 2

This suggests that the empirical formula is probably CH2. The empirical formula mass is therefore approximately 14 units (12 + 1+ 1).

168g mol−1 is the molar mass (rounded to a whole number) and 168 / 14 = 12

So there are 12 empirical formula units (of CH2) in the molecular formula.

Therefore the molecular formula is C12H24.

(Once the approximate 1:2 ratio of C:H is known, the last part can also be done by trial and error: 12×12 = 144, 144+24 = 168)

15

Complete this question without a calculator.

What is the percentage yield (%) when 14 g of ethene produces 12 g of ethanol?

C2H4 + H2O → C2H5OH

Using Mr(ethene) = 28 and Mr(ethanol) = 46.

Percentage yield = (experimental yield / theoretical yield) × 100%

Moles = mass / molar mass

Moles of ethene = 14 / 28

Moles of ethanol = 12 / 46

The reaction has a 1:1 molar ratio as seen in the equation. So the theoretical yield of ethanol in moles will be the same as the moles of ethene.

Therefore the percentage yield will be: (moles of ethanol / moles of ethene) × 100%

= (12 / 46) / (14 / 28) × 100%

(The easiest way to deal with this is probably to remember that dividing by a fraction is the same as multiplying by the inverted (upside down) fraction. So...)

= $${12\over 46}$$ × $${28\over 14}$$ × 100%

= $${12\times28\times100 \over 14\times46}$$

16

Complete this question without a calculator.

What is the value of x when 36.0g of MnCl2.xH2O is heated to dryness leaving 31.5g of anhydrous MnCl2?

Using Ar: Mn = 55, Cl = 35.5, H = 1, O = 16

The value for the water of crystallisation, x, can be found by finding the integer ratio of moles of MnCl2 : moles of H2O. (x is always given as a whole number.)

Moles (mol) = mass (g) / molar mass (g mol−1)

The moles of MnCl2 can be found using the mass of the anhydrous salt: 31.5 / 126 = $${1\over 4}$$ = 0.25 mol

The mass of water can be found by subtracting the mass of anhydrous salt from hydrated salt: 36.0 − 31.5 = 4.5 g

And moles of water = 4.5 / 18 = $${1\over 4}$$ = 0.25 mol

The ratio of moles of MnCl2 : moles of H2O is therefore 0.25 : 0.25 and the integer ratio is therefore 1 : 1.

The value of x is therefore 1.