For the following reaction, 0.689 grams of hydrogen gas are allowed to react with 12.8 grams of ethylene (C2H4) .

Question

hydrogen(g) + ethylene (C2H4)(g) ethane (C2H6)(g)

What is the maximum mass of ethane (C2H6) that can be formed? grams

What is the FORMULA for the limiting reagent? 

What mass of the excess reagent remains after the reaction is complete? grams

Answer 1

Again, ALWAYS start by writing the balanced equation: H₂(g) + C₂H₄(g) ==> C₂H₆(g)

Next, find moles of each reactant that is present:

>> moles H₂ gas = 0.689 g H₂ x 1 mole/2 g = 0.345 moles H₂ gas

>> moles of C₂H₄ gas = 12.8 g C₂H₄ x 1 mole/28 g = 0.457 moles C₂H₄ gas

Next, find which reactant is limiting.  This is most easily done by dividing moles of each gas by the corresponding coefficient in the balanced equation. Thus, H₂ = 0.345/1 = 0.345 and for C₂H₄ you have 0.457/1 = 0.457.  0.345 is the smallest, so H₂ gas is limiting.

To find maximum amount of C₂H₆ formed, this will now be determined by the amount of limiting reactant H₂ gas.

0.345 moles H₂ x 1 mole C₂H₆/mole H₂ = 0.345 moles C₂H₆

Mass C₂H₆ = 0.345 moles x 30 g/mole = 10.4 grams C₂H₆ is maximum amount formed

Formula of limiting reagent is H₂

Mass of excess reagent (C₂H₄) = 0.345 moles H₂ x 1 mole C₂H₄/mole H₂ = 0.345 moles C₂H₄ used and since you started with 0.457 moles, the moles left over = 0.457 - 0.345 = 0.112 moles C₂H₄

Mass = 0.112 moles x 28 g/mole = 3.14 g of C₂H₄ left over.