Ethane is formed from the reaction carbon and hydrogen:

2C(s) + 3 H2(g) C2H6(g)

If 7.08 g of carbon and 4.92 g of hydrogen are allowed to react, how much of the excess reactant (the “non-limiting reactant”) is left over?

a. 1.78 g
b. 2.16 g
c. 2.78 g
d. 3.14 g
e. 4.92 g

H2 is in excess. There will remain 3.14 grams of H2. Option D is correct

Explanation:

step 1: Data given

Mass of Carbon = 7.08 grams

Mass of hydrogen = 4.92 grams

Molar mass carbon = 12.01 g/mol

Molar mass hydrogen (H2) = 2.02 g/mol

Step 2: The balanced equation

2C(s) + 3 H2(g) → C2H6(g)

Step 3: Calculate moles carbon

Moles carbon = mass carbon / moalr mass carbon

Moles carbon = 7.08 grams / 12.01 g/mol

Moles carbon = 0.590 moles

Step 4: Calculate moles H2

Moles H2 = 4.92 grams / 2.02 g/mol

Moles H2 = 2.44 moles

Step 5: Calculate limiting reactant

For 2 moles carbon we nee 3 moles hydrogen to produce 1 mol C2H6

Carbon is the limiting reactant. It will completely be consumed  (0.590 moles). H2 is in excess. There will react 3/2 * 0.590 = 0.885 moles H2

There will remain 2.44 - 0.885 = 1.555 moles H2

Step 6: Calculate mass H2 remaining

Mass H2 = 1.555 moles * 2.02 g/mol

Mass H2 = 3.14 grams

H2 is in excess. There will remain 3.14 grams of H2. Option D is correct