Answer to Question #107351 in General Chemistry for Jessica Godinez

Question

Calculate the mass (in grams) of H2O produced from the reaction of 56.20 g C4H10 with 85.30 g O2.

C4H10 (g) + O2 (g) → CO2 (g) + H2O (l)

Answer 1

C₄H₁₀(g)  + 6,5O₂(g) → 4CO₂(g) + 5H₂O(l)

n(C₄H₁₀)=m(C₄H₁₀)/M(C₄H₁₀)=56.20/58=0.969mol

n(O₂)=m(O₂)/M(O₂)=85.30/32=2.666mol

n(C4H10)/1 and n(O2)/6.5

0.969/1 and 2.666/6.5

0.969 and 0.43

The mass of water is calculated by the deficiency, by n(O₂)

n(O₂)/n(H₂O)=6.5/5

2.666*5=6.5*n(H₂O)

n(H2O)=2.05mol

V(H2O)=n(H2O)*V_m=2.05*22.4=45.92l=45920ml

ρ(H2O)=1g/ml

m(H2O)=ρ*V(H2O)=1*45920=45920g

Answer - 45920g