Question

What mass of glucose is required to produce $88\text{g}$ of $C{O}_2$ on complete combustion?

• A. $30\text{g}$
• B. $45\text{g}$
• C. $60\text{g}$
• D. $22\text{g}$

Answer 1

The correct option is C $60\text{g}$

$C_6{H}_{12}{O}_6+6O_2\to 6C{O}_2+6H_{2}O$
Gram molar mass of $C_6{H}_{12}{O}_6=(6\times 12)+(12\times 1)+(6\times 16)=180\text{g}$
Now,
$(6\times 44)\text{g}$ of $C{O}_2$ is produced by $180\text{g}$ of glucose.
So, $88\text{g}$ of $C{O}_2$ is produced by $(\frac{180}{6\times 44}\times 88)=60\text{g}$ of glucose.