Question

Mass of sucrose $C_{12}{H}_{22}{O}_{11}$ produced by mixing 84 g of carbon, 12 g of hydrogen and 56 lit.$O_2$ at 1atm and 273 K according to given reaction is

C(s)+$H_2$+$O_2$$⟶$$C_{12}{H}_{22}{O}_{11}$

• A. 138.5
• B. 155.5
• C. 172.5
• D. 1499.5

Answer 1

The correct option is B

155.5

12C +11$H_2$+$\frac{11}{2}$$O_2$$⟶$$C_{12}{H}_{22}{O}_{11}$

7 moles 6 moles 2.5moles

$\frac{n_2}{\frac{11}{2}}$ =$n_{\left(C_{1}2H_{22}{O}_{11}\right)}$

$\frac{2.5}{\frac{11}{2}}$ =$n_{\left(C_{1}2H_{22}{O}_{11}\right)}$

$n_{\left(C_{1}2H_{22}{O}_{11}\right)}$ =$\frac{5}{11}$

$n_{\left(C_{1}2H_{22}{O}_{11}\right)}$ =$\frac{5}{11}$$\times$342 =155.5gm