Question

The combustion of a sample of butane C4H10 produced 2.46 grams of water? 2C4H10 + 13O2 ====== 8CO2 + 10H2O
A-How many moles of water formed?
B-How many moles of butane burned?
C-How many grams of butane burned?
D-How many oxygen was used up in moles? in grams? ​​

Answer 1

Dear Student,

$$\begin{gathered} 2{\mathrm{C}}_4{\mathrm{H}}_{10}+13{\mathrm{O}}_2\to 8{\mathrm{CO}}_2+10{\mathrm{H}}_2\mathrm{O} \\ 2\mathrm{mol}13\mathrm{mol}8\mathrm{mol}10\mathrm{mol} \\ \mathrm{A}.\mathrm{Moles}\mathrm{of}\mathrm{water}\mathrm{formed}=\frac{\mathrm{Mass}}{\mathrm{Molar}\mathrm{mass}}=\frac{2.46}{18}=0.1366\mathrm{mol} \\ \mathrm{B}.2\mathrm{mol}\mathrm{of}{\mathrm{C}}_4{\mathrm{H}}_{10}\mathrm{gives}10\mathrm{mol}\mathrm{of}{\mathrm{H}}_2\mathrm{O} \\ \mathrm{So},0.1366\mathrm{mol}\mathrm{of}{\mathrm{H}}_2\mathrm{O}\mathrm{is}\mathrm{given}\mathrm{by}\frac{2}{10}\times 0.1366=0.0273\mathrm{mol}\mathrm{of}\mathrm{butane} \\ \mathrm{C}.\mathrm{Mass}\mathrm{of}\mathrm{butane}\mathrm{formed}=\mathrm{Moles}\times \mathrm{Molar}\mathrm{mass}=0.0273\times 58=1.584\mathrm{g} \\ \mathrm{D}.2\mathrm{mol}\mathrm{of}\mathrm{butane}\mathrm{uses}13\mathrm{mol}\mathrm{of}{\mathrm{O}}_2 \\ \mathrm{So},0.0273\mathrm{mol}\mathrm{of}\mathrm{butane}\mathrm{uses}\frac{13}{2}\times 0.0273=0.1774\mathrm{mol}\mathrm{of}{\mathrm{O}}_2 \\ \mathrm{and},\mathrm{mass}\mathrm{of}{\mathrm{O}}_2=\mathrm{moles}\times \mathrm{molar}\mathrm{mass}=0.1774\times 32=5.678\mathrm{g} \end{gathered}$$