$\begin{array}{ccccc} 2\mbox{H}_{2} & + & \mbox{O}_{2} & \rightarrow & 2\mbox{H}_{2}\mbox{O}\\ 2 & : & 1 & : & 2 \end{array}$

mol $\mbox{H}_{2}=1$ mol

$\begin{aligned}M_{r}\mbox{ H}_{2}\mbox{O} & =\left(2\times A_{r}\mbox{ H}\right)+\left(1\times A_{r}\mbox{ O}\right)\\ & =(2\times1)+(1\times16)\\ & =2+16\\ & =18 \end{aligned}$

$\begin{aligned}\mbox{mol H}_{2}\mbox{O} & =\frac{2}{2}\times\mbox{mol H}_{2}\\ & =\frac{2}{2}\times1\mbox{ mol}\\ & =1\mbox{ mol} \end{aligned}$

$\begin{aligned}\mbox{massa H}_{2}\mbox{O}\mbox{teori} & =\mbox{mol H}_{2}\mbox{O}\times M_{r}\mbox{ H}_{2}\mbox{O}\\ & =1\times18\mbox{ g}\\ & =18\mbox{ g} \end{aligned}$

$\begin{aligned}\mbox{persentase hasil} & =\frac{\mbox{massa sebenarnya}}{\mbox{massa teoritis}}\times100\%\\ & =\frac{15}{18}\times100\%\\ & =83,33\%. \end{aligned}$

$\begin{alignedat}{1}n_{\mbox{CaCO}_{3}} & =\frac{10}{100}\mbox{ mol}\\ & =0,1\mbox{ mol}\\ n_{\mbox{CaO}} & =\frac{1}{1}\times0,1\mbox{ mol}\\ & =0,1\mbox{ mol}\\ m_{\mbox{CaO}} & =0,1\mbox{ mol}\times(40+16)\mbox{g/mol}\\ & =5,6\mbox{ g}\\ \%\mbox{hasil} & =\frac{\mbox{m yg didapat}}{\mbox{m teori}}\times100\%\\ & =\frac{4}{5,6}\times100\%\\ & =71,4\%. \end{alignedat}$

$\begin{array}{c} 2\mbox{N}_{2}\mbox{O}_{3}(g)\\ 2\\ 16\mbox{ L}\\ \\ \end{array}$ $\begin{array}{cc} \rightarrow & \mbox{N}_{2}\mbox{O}_{4}(g)\\ : & 1\\ & \frac{1}{2}\times16\mbox{ L}\\ & =8\mbox{ L} \end{array}$ $\begin{array}{cc} + & 2\mbox{NO(g)}\\ : & 2\\ \,^{\,^{\,}}\\ \,^{\,^{\,}} \end{array}$

$\begin{alignedat}{1}V_{\mbox{teori}} & =8\mbox{ L}\\ \%\mbox{hasil} & =\frac{V_{\mbox{aktual}}}{V_{\mbox{teori}}}\times100\%\\ 75\% & =\frac{V_{\mbox{aktual}}}{8\mbox{ L}}\times100\%\\ V_{\mbox{aktual}} & =6\mbox{ L}. \end{alignedat}$

$\begin{array}{ccccc} 2\mbox{SO}_{2} & + & \mbox{O}_{2} & \rightarrow & 2\mbox{SO}_{3}\\ & & 1,2\mbox{ L} & & \frac{2}{1}\times1,2\mbox{ L}\\ & & & & =2,4\mbox{ L} \end{array}$

$\begin{alignedat}{1}\%\mbox{hasil} & =\frac{2}{2,4}\times100\%\\ & =83,3\%. \end{alignedat}$

$\begin{alignedat}{1}V_{\mbox{aktual}} & =4\mbox{ L}\\ \%\mbox{hasil} & =\frac{V_{\mbox{aktual}}}{V_{\mbox{teori}}}\times100\%\\ 80\% & =\frac{4\mbox{ L}}{V_{\mbox{teori}}}\times100\%\\ V_{\mbox{teori}} & =5\mbox{ L}\\ V_{\mbox{hilang}} & =V_{\mbox{teori}}-V_{\mbox{aktual}}\\ & =5-4\mbox{ L}\\ & =1\mbox{ L.} \end{alignedat}$

$\begin{array}{c} \mbox{2SO}_{3}(g)\\ 5\mbox{ L}\\ \\ \end{array}$ $\begin{array}{cc} \rightarrow & 2\mbox{SO}_{2}(g)\\ & \frac{2}{2}\times5\mbox{ L}\\ & 5\mbox{ L} \end{array}$ $\begin{array}{cc} + & \mbox{O}_{2}(g)\\ & \frac{1}{2}\times5\mbox{ L}\\ & =2,5\mbox{ L} \end{array}$

$\begin{alignedat}{1}\%\mbox{hasil} & =\frac{2}{2,5}\times100\%\\ & =80\%\\ V_{\mbox{aktual SO}_{2}} & =80\%\times5\mbox{ L}\\ & =4\mbox{ L.} \end{alignedat}$

$\begin{alignedat}{1}\%\mbox{hasil} & =\frac{m_{\mbox{aktual}}}{m_{\mbox{teori}}}\times100\%\\ 80\% & =\frac{2,6\mbox{ gram}}{m_{\mbox{teori}}}\times100\%\\ m_{\mbox{teori}} & =3,25\mbox{ gram} \end{alignedat}$

massa teori $\mbox{C}_{2}\mbox{H}_{2}$= 3,25 gram.

mol teori $\mbox{C}_{2}\mbox{H}_{2}$$=\frac{3,25}{(24+2)}$$=0,125$ mol.

$\begin{array}{c} \mbox{CaC}_{2}\\ \frac{1}{1}\times0,125\mbox{ mol}\\ =0,125\mbox{ mol} \end{array}$ $\begin{array}{cc} + & 2\mbox{H}_{2}\mbox{O}\\ \\ \\ \end{array}$ $\begin{array}{cc} \rightarrow & \mbox{Ca(OH)}_{2}\\ \\ \\ \end{array}$ $\begin{array}{cc} + & \mbox{C}_{2}\mbox{H}_{2}\\ & 0,125\mbox{ mol}\\ \\ \end{array}$

$\begin{aligned}\mbox{m CaC}_{2} & =0,125\mbox{ mol}\times(40+24)\mbox{g/mol}\\ & =8\mbox{ gram}. \end{aligned}$

$\begin{aligned}V_{\mbox{teori}}-V_{\mbox{aktual}} & =2\mbox{ L}\\ V_{\mbox{teori}} & =V_{\mbox{aktual}}+2\mbox{ L} \end{aligned}$

$\begin{aligned}\%\mbox{hasil} & =\frac{V_{\mbox{aktual}}}{V_{\mbox{teori}}}\times100\%\\ 75\% & =\frac{V_{\mbox{aktual}}}{V_{\mbox{aktual}}+2\mbox{ L}}\times100\%\\ 75\%(V_{\mbox{aktual}}+2\mbox{ L}) & =V_{\mbox{aktual}}\times100\%\\ V_{\mbox{aktual}} & =6\mbox{ L} \end{aligned}$

$\begin{array}{c} 2\mbox{Na}(s)\\ 5,75\mbox{ gram}\\ \frac{5,75\mbox{ gram}}{23}\\ =0,25\mbox{ mol}\\ \\ 2\mbox{K}(s)\\ (15,5-5,75)\mbox{ gram}\\ \frac{9,75}{39}\mbox{ mol}\\ 0,25\mbox{ mol}\\ \\ \end{array}$ $\begin{array}{cc} + & 2\mbox{H}_{2}\mbox{O}(\ell)\\ \,^{\,^{\,}}\\ \,^{\,^{\,}}\\ \,^{\,^{\,}}\\ \,^{\,^{\,}}\\ + & 2\mbox{H}_{2}\mbox{O}(\ell)\\ \,^{\,^{\,}}\\ \,^{\,^{\,}}\\ \,^{\,^{\,}}\\ \\ \end{array}$ $\begin{array}{cc} \rightarrow & 2\mbox{NaOH}(aq)\\ \,^{\,^{\,}}\\ \,^{\,^{\,}}\\ \,^{\,^{\,}}\\ \,^{\,^{\,}}\\ \rightarrow & 2\mbox{KOH}(aq)\\ \,^{\,^{\,}}\\ \,^{\,^{\,}}\\ \,^{\,^{\,}}\\ \\ \end{array}$ $\begin{array}{cc} + & \mbox{H}_{2}(g)\\ \\ \\ & \frac{1}{2}\times0,25\mbox{ mol}\\ & 0,125\mbox{ mol}\\ + & \mbox{H}_{2}(g)\\ \\ \\ \,^{\,^{\,}} & \frac{1}{2}\times0,25\mbox{ mol}\\ & 0,125\mbox{ mol} \end{array}$

$\begin{aligned}n_{\mbox{total H}_{2}} & =0,125\mbox{ mol}+0,125\mbox{ mol}\\ & =0,25\mbox{ mol} \end{aligned}$

$\begin{alignedat}{1}V_{\mbox{total H}_{2}} & =n\times22,4\mbox{ L}\\ & =0,25\times22,4\mbox{ L}\\ & =5,6\mbox{ L}\\ \%\mbox{hasil} & =\frac{4}{5,6}\times100\%\\ & =71,4\%. \end{alignedat}$

$\begin{array}{cc} \mbox{CH}_{4} & +2\mbox{O}_{2}\\ 0,2\mbox{ mol}\\ \\ \end{array}$$\begin{array}{cc} \rightarrow & \mbox{CO}_{2}\\ & \frac{1}{1}\times0,2\mbox{ mol}\\ & =0,2\mbox{ mol} \end{array}$$\begin{array}{cc} + & 2\mbox{H}_{2}\mbox{O}\\ \\ \\ \end{array}$

$\begin{alignedat}{1}m_{\mbox{CO}_{2}} & =0,2\times44\mbox{ gram}\\ & =8,8\mbox{ gram}\\ \%\mbox{hasil} & =\frac{7,8}{8,8}\times100\%\\ & =88,6\% \end{alignedat}$

$\begin{array}{cc} \mbox{C}_{2}\mbox{H}_{4} & +\mbox{3O}_{2}\\ 0,2\mbox{ mol}\\ \\ \end{array}$$\begin{array}{cc} \rightarrow & 2\mbox{CO}_{2}\\ & \frac{2}{1}\times0,2\mbox{ mol}\\ & =0,4\mbox{ mol} \end{array}$$\begin{array}{cc} + & 2\mbox{H}_{2}\mbox{O}\\ \\ \\ \end{array}$

$\begin{alignedat}{1}m_{\mbox{CO}_{2}} & =0,4\times44\mbox{ gram}\\ & =17,6\mbox{ gram}\\ \%\mbox{hasil} & =\frac{15,4}{17,6}\times100\%\\ & =87,5\%. \end{alignedat}$

$\begin{array}{c} \mbox{2NH}_{3}(g)\\ 3,4\mbox{ gram}\\ n=\frac{3,4}{17}\mbox{mol}\\ =0,2\mbox{ mol} \end{array}$ $\begin{array}{cc} \rightarrow & \mbox{3H}_{2}(g)\\ \\ \,^{\,^{\,}}\\ \,^{\,^{\,}} \end{array}$ $\begin{array}{cc} + & \mbox{N}_{2}(g)\\ \\ \,^{\,^{\,}}\\ \,^{\,^{\,}} \end{array}$

$\begin{alignedat}{1}n_{\mbox{N}_{2}} & =\frac{1}{2}\times0,2\mbox{ mol}\\ & =0,1\mbox{ mol}\\ m_{\mbox{N}_{2}} & =0,1\times28\mbox{ gram}\\ & =2,8\mbox{ gram}\\ \%\mbox{hasil} & =\frac{2,1}{2,8}\times100\%\\ & =75\%\\ n_{\mbox{H}_{2}} & =\frac{3}{2}\times0,2\mbox{ mol}\\ & =0,3\mbox{ mol}\\ V_{\mbox{teori H}_{2}} & =0,3\times24\mbox{ dm}^{3}\\ & =7,2\mbox{ dm}^{3}\\ V_{\mbox{aktual H}_{2}} & =75\%\times7,2\mbox{ dm}^{3}\\ & =5,4\mbox{ dm}^{3}. \end{alignedat}$

$\begin{alignedat}{1}m_{\mbox{teori}}-m_{\mbox{aktual}} & =6\mbox{ gram}\\ m_{\mbox{teori}} & =6+m_{\mbox{aktual}}\\ n_{\mbox{teori}} & =\frac{6+m_{\mbox{aktual}}}{32}\\ n_{\mbox{NH}_{3}} & =\frac{2}{1}\times\frac{6+m_{\mbox{aktual}}}{32}\\ & =\frac{6+m_{\mbox{aktual}}}{16}\\ m_{\mbox{NH}_{3}} & =n\times M_{r}\\ 51\mbox{ gram} & =\frac{6+m_{\mbox{aktual}}}{16}\times17\\ m_{\mbox{aktual}} & =48\mbox{ gram}\\ \%\mbox{hasil} & =\frac{m_{\mbox{aktual}}}{m_{\mbox{teori}}}\times100\%\\ & =\frac{48}{48+6}\times100\%\\ & =88,9\%. \end{alignedat}$

$\begin{alignedat}{1}\%\mbox{hasil} & =\frac{V_{\mbox{aktual}}}{V_{\mbox{teori}}}\times100\%\\ 88,3\% & =\frac{800}{V_{\mbox{teori}}}\times100\%\\ V_{\mbox{teori}} & =960\mbox{ cm}^{3}\\ n_{\mbox{H}_{2}} & =\frac{960}{24000}\mbox{mol}\\ & =0,04\mbox{ mol}\\ \\ \end{alignedat}$

$\begin{array}{cc} \mbox{Zn} & +2\mbox{HCl}\\ \frac{1}{1}\times0,04\mbox{ mol}\\ =0,04\mbox{ mol} \end{array}$$\begin{array}{cc} \rightarrow & \mbox{ZnCl}_{2}\\ \\ \\ \end{array}$$\begin{array}{cc} + & \mbox{H}_{2}\\ & 0,04\mbox{ mol}\\ \\ \end{array}$

$\begin{alignedat}{1}m_{\mbox{Zn}} & =n\times A_{r}\mbox{ Zn}\\ & =0,04\mbox{ mol}\times65\mbox{g/mol}\\ & =2,6\mbox{ g}\\ m_{\mbox{Cu}} & =5-m_{\mbox{Zn}}\\ & =5-2,6\mbox{ g}\\ & =2,4\mbox{ g}. \end{alignedat}$

$\begin{array}{c} \mbox{C}_{2}\mbox{H}_{4}\\ 1,3\mbox{ gram}\\ \frac{1,3}{26}\mbox{mol}\\ 0,05\mbox{ mol}\\ \\ \end{array}$$\begin{array}{cc} + & 3\mbox{O}_{2}\\ \\ \\ \,^{\,^{\,}}\\ \,^{\,^{\,}} \end{array}$$\begin{array}{cc} \rightarrow & 2\mbox{CO}_{2}\\ \\ \,^{\,}\\ & \frac{2}{1}\times0,05\mbox{ mol}\\ & =0,1\mbox{ mol} \end{array}$$\begin{array}{cc} + & 2\mbox{H}_{2}\mbox{O}\\ \\ \\ \,^{\,^{\,}}\\ \,^{\,^{\,}} \end{array}$

$\begin{alignedat}{1}V_{\mbox{aktual CO}_{2}} & =0,1\times44\mbox{ g}\\ & =4,4\mbox{ g}\\ \%\mbox{hasil} & =\frac{3,3\mbox{ gram}}{4,4\mbox{ gram}}\times100\%\\ & =75\% \end{alignedat}$

Maka, untuk pembakaran kedua:

$\begin{aligned}n\mbox{ CO}_{2}\mbox{aktual} & =22,4\times\frac{1}{22,4}\mbox{ mol}\\ & =1\mbox{ mol} \end{aligned}$

$\begin{aligned}75\% & =\frac{n\mbox{ CO}_{2}\mbox{aktual}}{n\mbox{ CO}_{2}\mbox{teoritis}}\times100\%\\ n\mbox{ CO}_{2}\mbox{teoritis} & =1,33\mbox{ mol} \end{aligned}$

$\begin{array}{c} \mbox{C}_{2}\mbox{H}_{4}\\ \\ \,^{\,^{\,}}\\ \\ \,^{\,^{\,}} \end{array}$$\begin{array}{cc} + & 3\mbox{O}_{2}\\ \\ \\ & \frac{3}{2}\times1,33\mbox{ mol}\\ & =\mbox{2 mol} \end{array}$$\begin{array}{cc} \rightarrow & 2\mbox{CO}_{2}\\ & 1,33\mbox{ mol}\\ \\ \\ \\ \end{array}$$\begin{array}{cc} + & 2\mbox{H}_{2}\mbox{O}\\ \\ \,^{\,^{\,}}\\ \,^{\,^{\,}}\\ \\ \end{array}$

$\begin{alignedat}{1}V_{\mbox{O}_{2}} & =2\times22,4\mbox{L}\\ & =44,8\mbox{ L}. \end{alignedat}$

Misal mol $\mbox{Cu}\left(\mbox{NO}_{3}\right)_{2}$$=2x\mbox{ mol}$

$\begin{array}{c} \mbox{Cu}\left(\mbox{NO}_{3}\right)_{2}(s)\\ 2x\mbox{ mol} \end{array}$$\begin{array}{cc} \rightarrow & 2\mbox{CuO}(s)\\ \\ \end{array}$$\begin{array}{cc} + & 4\mbox{NO}_{2}(g)\\ & 4x\mbox{ mol} \end{array}$$\begin{array}{cc} + & \mbox{O}_{2}(g)\\ & x\mbox{ mol} \end{array}$

$\begin{alignedat}{1}n_{\mbox{NO}_{2}} & =4x\mbox{ mol}\\ m_{\mbox{NO}_{2}} & =4x\times(14+16+16)\mbox{gram}\\ & =184x\mbox{ gram}\\ n_{\mbox{O}_{2}} & =x\mbox{ mol}\\ m_{\mbox{O}_{2}} & =x\times32\mbox{gram}\\ & =32x\mbox{ gram}\\ m_{\mbox{total gas}} & =184x+32x\\ & =216x\mbox{ gram} \end{alignedat}$

$\begin{alignedat}{1}\%\mbox{hasil} & =\frac{m_{\mbox{aktual}}}{m_{\mbox{teori}}}\times100\%\\ 94,4\% & =\frac{20,4}{216x}\times100\%\\ x & =0,1\\ n_{\mbox{Cu}\left(\mbox{NO}_{3}\right)_{2}} & =0,2\mbox{ mol}\\ m_{\mbox{Cu}\left(\mbox{NO}_{3}\right)_{2}} & =37,5\mbox{ gram}. \end{alignedat}$