Stoichiometry

Reading Equations using MolesCa + H2SO4 CaSO4 + H2

The above reaction reads:

1 amount of Ca reacts with 1 amount of H2SO4 to give 1 amount of CaSO4 and 1 amount of H2 gas.

(we don’t usually put in the ones – but if we did it would look like this)

The amount we use is the mole

Now the reaction reads:

1 mole of Ca reacts with 1 mole of H2SO4 to give 1 mole of CaSO4 and 1 mole of H2 gas.

1 1 1 1

Reading Equations using MolesCa + H2SO4 CaSO4 + H2

Questions:

1. If we have 1 mole of Ca how many moles of H2SO4 are needed to fully react the Ca?

2. If we had 0.5 mole of Ca how many moles of H2SO4 would be required?

3. If 0.5 mole of Ca reacted with 0.5 mole of H2SO4 how many moles of CaSO4 and how many moles of H2 gas are formed?

Ans: 1mole

Ans: 0.5 mole

Ans : 0.5 mole of H2 gas, 0.5 mole of CaSO4

Reading Equations using MolesCa + H2SO4 CaSO4 + H2

Questions:

If 0.5 mole of Ca reacted with 0.5 mole of H2SO4 how many moles of CaSO4 and how many moles of H2 gas are formed?

Ans : 0.5 mole of H2 gas, 0.5 mole of CaSO4

Given the Ar (atomic mass) values

H = 1 gmol-1 Ca = 40 gmol-1 S = 32 gmol-1 O =16 gmol-1

Find the molar masses (Mr) of H2SO4 , CaSO4 and H2

Then use these and the mole (n) formula to work out the mass of H2 gas and the mass of CaSO4 formed

(Hint you will need to rearrange the formula)

Reading Equations using Moles

Fe2O3 + 3CO 2Fe + 3CO2

Q1. If 200g of Fe2O3 is reacted in the above equation what mass of Fe is produced? (Mr Fe2O3 = 160 gmol-1, Mr Fe = 55.85 gmol-1)

321

3232

0Fe moles 25.1 gmol160

g200

)O(Fe Mrm

0Fe of moles

From the equation we see that for every mole of Fe2O3 that reacts 2 moles of Fe are formed)

This means that if we can find the number of moles of Fe2O3 we can then work out the number of moles of Fe and then the mass of Fe

Using the mole ratio from the equation how many moles of Fe are formed? (unknown/known)

You got it 2.5 moles of Fe is formed –now work out the mass of Fe

Reaction ratio

Mg + 2 HCl MgCl2 + H2

Work out how many moles of HCl is used to completely react 60grams of Mg metal in the following reaction? Draw the grid

2

Mrm

n

mass

Mg

?

60.0 grams

1-24gmol60g

n = 2.5 moles

24.0 gmol -1Mr

HCl

36.5 gmol -1

1 Therefore (unknown/known) 2mol / 1mol x moles of Mg = moles of HCl

Therefore 2 x 2.5 moles = 5 moles of HCl

5.00 moles

183 grams

What was the mass of HCl used?

Mrm

n

m = 5 x Mr = 5 mol x 36.5 gmol-1

= 182.5 g

= 183 g (3sf)