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)

Md Ariful Islam

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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)*Vm=2.05*22.4=45.92l=45920ml

ρ(H2O)=1g/ml

m(H2O)=ρ*V(H2O)=1*45920=45920g

Answer - 45920g

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