EXERCISES 1. What mass of copper can be made by reducing 16g of copper oxide wit

Question

EXERCISES 1. What mass of copper can be made by reducing 16g of copper oxide with hyvdrogen? Equation Masses Mr Mass of Cu CuO +H2 Cu+H2O 168 80 64 What mass of Magnesium will react with 19.6 g of sulphuric acid Equation Mg + H2S04- MgSO4+ H2 Masses Mr Mass of Mg x 19.68 3. What mass of iron can be made from 40g of iron oxide? Equation Masses Mr Mass of Fe Fe203 +3CO-2Fe + 3C02 What mass of carbon can be made from 18g of glucose? C&H120;% 6C +6H20 Equation Masses Mr Mass of C 5. What mass of Hydrogen is needed to make 50g of ammonia? Equation Masses Mr Mass of NaCl N2 +3H2-2NHs

Explanation / Answer

1) the equation is

CuO + H2 --> Cu(s) + H2O

Moles of CuO used = Mass / Molar mass = 16 / 80 = 0.2 moles

so moles of copper obtained = Moles of CuO reduced = 0.2

Mass of Cu = Moles X atomic mass = 0.2 X 64 = 12.8g

2) Reaction is

Mg + H2SO4 ---> MgSO4 + H2

Each mole of Mg will react with one mole of H2SO4,

Moles of H2SO4 reacted = Mass / molar mass = 19.6 / 98 = 0.2 moles

Moles of Mg reacted = 0.2 moles

Mass of Mg reacted = Moles X atomic mass = 0.2 X 24 = 12 g

3) Reaction is

Fe2O3 + 3CO ---> 2Fe + 3CO2

So from each mole of Fe2O3, two moles of Fe will be obtained

Molar mass of Fe2O3 = 160 g / mole

Atomic weight of Fe = 56g / mole

Moles of Fe2O3 used = Mass / Molar mass = 40g / 160 = 0.25 moles

Moles of Fe obtained = 2 X 0.25 = 0.5 moles

Mass of Fe = Moles X atomic weight = 0.5 X 56 = 28 g

4) Reaction is

C6H12O6 ---> 6C + 6H2O

from each moles of glucose six moles of carbon will be obtained

Molar mass of glucose = 180g /mole

Atomic mass of carbon = 12g /mole

Moles of glucose reacted = Mass / Molar mass = 18 /180 = 0.1 mole

Moles of carbon obtained = 6 X 0.1 = 0.6 moles

Mass of carbon obtained = Moles X atomic mass = 0.6 X 12 = 7.2 grams

5) the reaction is

N2 + 3H2 ---> 2NH3

as per equation three moles of hydrogen will give two moles of NH3

molar mass of ammonia = 17 g / mole

atomic mass of H2 = 2g /mole

Moles of ammonia formed = Mass / Molar mass = 50 / 17 = 2.9 moles

Moles of hydrogen needed = 1. 5 X 2.9 = 4.35 moles

Mass of hydrogen required = Moles X atomic mass = 4.35 X 2 = 8.7 grams

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