Question

The reaction for the combustion of octane (C8H18) is given in this reaction:

2C8H18 + 25O2 → 16CO2 + 18H2O

How many grams of H2O are produced if 228 grams of octane are burned?

181 grams

228 grams

81 grams

324 grams

Answer 1

\(2C_8H_{18}~+~25O_2\) ------->\(16CO_2~+~18H_2O\)

here we see that 2 moles of \(C_8H_{18}\) (Octane) react to produce 18 moles of \(H_2O\)

so 228 grams of Octane are burned
moles of octane burned= \(\large \frac{mass}{molar~mass}\)
so to find moles we gotta know the molar mass

molar mass of Octane (\(C_8H_{18}\))= \(8 \times Molar~mass~of~C+18 \times Molar~mass~of~H\)

molar mass of C is 12 and of H is 1
so we get \(Molar~mass~of~C_8H_{18}=114\)

nw we have molar mass of octane and also the mass burned
so
moles of octane burned= \(\large \frac{mass}{molar~mass}\)
moles of octane burned= \(\large \frac{228}{114}\)
moles of octane burned= \(2\) moles

we know that 2 moles of octane give 18 moles of \(H_2O\)
so we get 18 moes of \(H_2O\)
mass of 18 moles of \(H_2O\) =\(moles \times Molar~mass\)
molar mass of \(H_2O\) is 18
so put it in
mass of 18 moles of \(H_2O\) =\(18\times 18\)

we get mass of \(H_2O\) used =324grams