Question

Gaseous butane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water. What is the theoretical yield of water formed from the reaction of 1.7 g of butane and 2.2 g of oxygen gas? Round your answer to 2 significant figures.

I thought it was 2.6g but it said it was wrong

Answer 1

C₄H₁₀ + 13/2 O₂ --> 4 CO₂ + 5 H₂O

1 mole butane produces 5 moles f gaseous water. 1.7 g of butane=1.7/58.12 mole=0.02925 mole (M.W butane=58.12 g/ mole).

2.2 g of oxygen = 2.2/18 mole=0.122 mole (M.W_O2=18 g/mole). From the above chemical equation one can say, 0.02925 mole of butane=13/2 *0.02925 mole of oxygen=0.19 mole of oxygen.

Therefore, 0.122 mole of oxygen reacts with 2*0.122 /13 mole of butane to produce 10*0.122/13 mole=0.09385 mole of gaseous water.