Question

For the following reaction, 5.28 grams of hydrogen gas are allowed to react with 9.91 grams of ethylene (c2h4) .

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

The reaction:

C2H4 + H2 = C2H6

what is the maximum mass of ethane (c2h6) that can be formed?

The amount of hydrogen:

n(H2) = m(H2) / M(H2) = 5.28 / 2 = 2.64 mol

The amount of ethylene:

n(C2H4) = m(C2H4) / M(C2H4) = 9.91 / 28 = 0.354 mol

We see that the hydrogen is an excess reagent

The maximum mas of ethane is:

m(C2H6) = n(C2H4) * M(C2H6) = 0.354 * 30 = 10.62 g

what is the formula for the limiting reagent? - C2H4

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

The amount of hydrogen after reaction:

n2(H2) = 2.64 - 0.354 = 2.286 mol

m2(H2) = 2.286 * 2 = 4.572 g

Answer: m(C2H6)=10.62 g; C2H4; m2(H2)=4.572 g