For the following reaction, 4.92 grams of hydrogen gas are allowed to react with 10.2 grams of ethylene (C2H4) .

Question

For the following reaction, 4.92 grams of hydrogen gas are allowed to react with 10.2 grams of ethylene.

hydrogen(g) + ethylene(g) ---> ethane (g)

What is the maximum mass of ethane 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

Write the correctly balanced equation...

H2(g) + C2H4(g) ==> C2H6(g)

Find which reactant, if either, is limiting...

4.92 g H2 x 1 mol H2/2 g = 2.46 moles H2 present

10.2 g C2H4 x 1 mol C2H4/28 g = 0.364 moles C2H4 present

Since these two reactants react in a 1:1 mole ratio, C2H4 will be limiting. We use this to find mass of ethane.

mass of ethane formed = 0.364 moles C2H4 x 1 mol C2H6/mol C2H4 x 30 g/mol C2H6 = 10.9 g formed

Limiting reagent = C2H4

To find mass of excess reagent (H2) remaining we see how much was used up and subtract that from how much we started with.

H2 used up: 0.364 moles C2H4 x 1 mol H2/mol C2H4 = 0.364 moles H2 used up.

We started with 2.46 moles of H2. So, 2.46 - 0.364 = 2.096 moles remaining

mass of H2 remaining = 2.096 moles H2 x 2 g/mol = 4.19 g H2 remaining (3 sig. figs.)