Question

when a mixture of 10.0g of acetylene, C2H2 and 10.0g of oxgyen, O2 is ignited the resulant combustion reaction procedures CO2 and H2O

2C2H2 + SO2= 4Co2 + 2H2O

what is limiting reactant?

what is the excess reactant?

how many grams of CO2 are produce after the reaction is complete?

how many grams of H2O are produced after the reaction is complete?

how many grams of excess reactant (C2H2 or O2) remain?

8.04 grams of CO2 were actally produce calculate the percentage yield.

Answer 1

2C₂H₂ + 5O₂ 4CO₂ + 2H₂O

10.0 g of acetylene (C₂H₂)

Molar mass of C₂H₂ = 26.04 g/mol

So, 10.0 g of C₂H₂ = (10.0 g) / (26.04 g/mol) = 0.38 mol

10.0g of oxgyen (O₂)

Molar mass of O₂= 16.0 g/mol

So, 10.0 g of C₂H₂ = (10.0 g) / (16.0 g/mol) = 0.625 mol

Here 2 moles of C₂H₂ reacts with 5 moles of O₂.

0.76 moles of C₂H₂ reacts with 3.125 moles of O₂.

Hence the limiting reagents is C₂H₂.

O₂ is the excess reagents, since it is used in excess moles.

2 moles of C₂H₂produces 4 moles of CO₂.

1 moles of C₂H₂produces 2 moles of CO₂.

0.38 moles of C₂H₂produces 0.76 moles of CO₂.

Molar mass of CO₂ = 44.01 g/mol.

So, 1 mole of CO₂ = 44.01 g

0.76 moles of CO₂ = (0.76 x 44.01) g = 33.45 g.

Here 2 moles of C₂H₂produces 2 moles of H₂O.

1 moles of C₂H₂produces 1 moles of H₂O.

0.38 moles of C₂H₂ produces 0.38 moles of H₂O.

Molar mass of H₂O = 18 g/mol.

So, 1 mole of H₂O= 18 g

0.38 moles of H₂O= (0.38 x 18) g = 6.84 g

0.38 mol of C₂H₂ reacts with 0.38 mol of O₂.

Excess mol of O₂= (0.625 - 0.38) mol = 0.245 mol

Excess grams of O₂= 0.245 mol x 16 g/mol = 3.92 g

Therotical yield = 33.45 g

Experimental yield = 8.04 g

Therefore %Yield = (8.04/33.45) x 100 = 24.04%