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Question

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42g/mol C2H6   32g/mol O2 44g/mol CO2 18g/mol H2O

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

193.2 g C2H6 reacts with 500g of oxygen.

1. What is the mass of CO2 produced? 30.07 u
2. What is the mass of H2O produced? 
3. What is the mass of the remaining reactant?470g
4. Is the Law of conservation of mass obeyed? (Dis)Prove it.

Answer 1

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O ... balanced equation

Find limiting reactant:

193.2 g C2H6 x 1 mol C2H6 / 42 g x 4 mol CO2 / 2 mol C2H6 x 44 g CO2/mol = 405 g CO2

500 g O2 x 1 mol O2 / 32 g x 4 mol CO2 / 7 mol O2 x 44 g CO2/mol = 393 g CO2

O2 is LIMITING

1) 393 g CO2 is formed

2) 15.6 mol O2 x 6 mol H2O / 7 mol O2 x 18 g H2O/mol = 241 g H2O

3) Remaining reactant = C2H6

Initial moles C2H6 = 193.2 g / 42 g/mol = 4.6 mol

moles used = 15.6 mol O2 x 2 mol C2H6 / 7 mol O2 = 4.46 mol used

mass C2H6 remaining = 4.60 mol - 4.46 mol = 0.14 moles x 42 g/mol = 5.9 g remain

4) mass of reactants = 193 g + 500 g = 693 g

mass of products + mass of remaining reactant = 393 g + 241 + 6 g = 640 g

This doesn't agree with the law of conservation of mass because the mass of staring material doesn't equal the mass of ending material. A good reason for this is because a gas (CO2) is formed and some of it may not have been collected and may have escaped into the atmosphere. Another possible reason is the calculations are in error so please do check them.