Oxyacetylene torches are used for welding at temperature close to 200°C.These temperatures are due to combustion of acetylene with oxygen.

Question

Oxyacetylene torches are used for welding at temperature close to 200°C.These temperatures are due to combustion of acetylene with oxygen.

C2H2 (g) + O2 (g) -> CO2 (g) + H2O (g)

a) Balance the above equation.

b) Starting with 150g of both C2H2 and 02 which reagent is limiting?

c) Calculate the theoretical yield of H2O from this mixture.

d) If 30.0 g of water forms, what is the percent yield?

e) How many grams of excess reactant remain unreacted?

Answer 1

@Mila C: You posted NINE questions all at the same time. It is not likely that any ONE person will answer all of them, and maybe no one will answer them. Did you even make an attempt to do them? If so, where are you having difficulty? We are not here to do your homework, but we are here to help you. I will do this one, and maybe one other, but you should really try them.

a). 2C₂H₂ (g) + 5O₂ (g) -> 4CO₂ (g) + 2H₂O (g) .. balanced equation

b). To find limiting reactant, divide moles of each by the corresponding coefficient in the balanced equation:

C₂H₂: 150 g x 1 mole C₂H₂ / 26.0 g = 5.77 mols (÷2->2.9)

O2: 150 g x 1 mole O₂ / 32 g = 4.69 mols (÷5->0.9)

0.9 is less than 2.9 so O₂ is the limiting reactant

c). Theoretical yield of H₂O:

4.69 mols O₂ x 2 mols H₂O / 5 mols O₂ x 18 g H₂O / mol H₂O = 33.8 g H₂O

d). % yield H₂O = actual yield / theoretical yield (x100%) = 30.0 g / 33.8 g (x100%) = 88.8%

e). To find grams excess C₂H₂, find amount used, and subtract from amount started with:

C₂H₂ used up: 4.69 mols O₂ x 2 mols C₂H₂ / 5 mols O₂ x 26 g C₂H₂ / mole C₂H₂ = 48.8 g C₂H₂ used

C₂H₂ remaining = 150 g - 48.8 g = 101 g remaining