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Question Number 131935 by Salman_Abir last updated on 09/Feb/21

Answered by physicstutes last updated on 10/Feb/21

$${2\mathrm{CH}}_3\mathrm{OH}+{3\mathrm{O}}_2\to 2{\mathrm{CO}}_2+4{\mathrm{H}}_2\mathrm{O}$$$$\mathrm{number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{CH}}_3\mathrm{OH}:$$$$\frac{6.40\cancel{\mathrm{g}{\mathrm{CH}}_3\mathrm{OH}}}{1}\times \frac{1\mathrm{mol}{\mathrm{CH}}_3\mathrm{OH}}{32\cancel{\mathrm{g}{\mathrm{CH}}_3\mathrm{OH}}}=0.2\mathrm{mol}{\mathrm{CH}}_3\mathrm{OH}$$$$\mathrm{mole}\mathrm{per}\mathrm{coefficient}\mathrm{ratio}=\frac{0.2}{2}=0.100$$$$\mathrm{number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{O}}_2:$$$$\frac{10.2\cancel{\mathrm{g}{\mathrm{O}}_2}}{1}\times \frac{1\mathrm{mol}{\mathrm{O}}_2}{32\cancel{\mathrm{g}{\mathrm{O}}_2}}=0.319\mathrm{mol}{\mathrm{O}}_2$$$$\mathrm{mole}\mathrm{per}\mathrm{coefficient}\mathrm{ratio}=\frac{0.319}{3}=0.106$$$$\Rightarrow {\mathrm{CH}}_3\mathrm{OH}\mathrm{is}\mathrm{the}\mathrm{limiting}\mathrm{reagent}.$$$$\mathrm{now}\frac{0.2\cancel{\mathrm{mol}{\mathrm{CH}}_3\mathrm{OH}}}{1}\times \frac{2\cancel{\mathrm{mol}{\mathrm{CO}}_2}}{2\cancel{\mathrm{mol}{\mathrm{CH}}_3\mathrm{OH}}}\times \frac{44\mathrm{g}{\mathrm{CO}}_2}{1\cancel{\mathrm{mol}{\mathrm{CO}}_2}}=8.8\mathrm{g}{\mathrm{CO}}_2$$$$\mathrm{percentage}\mathrm{yield}=\frac{\mathrm{actual}\mathrm{yield}}{\mathrm{theoritical}\mathrm{yield}}\times 100\mathrm{\%}=\frac{6.12\mathrm{g}}{8.80\mathrm{g}}\times 100$$$$\mathrm{percentage}\mathrm{yield}=69.5\mathrm{\%}\approx 70.00\mathrm{\%}$$

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