Question

when your body burns 1.00 g of sucrose, blank 1 grams of oxygen is burned, according to the following reaction, c12h22o11 + 12 o2 → 12co2 + 11 h2o. blank 1 add your answer

Explanation:

Step1: Determine molar mass of sucrose

The molar mass of $C_{12}H_{22}O_{11}$ ($M_{sucrose}$) is calculated as: $12\times12 + 22\times1+11\times16=342\ g/mol$.

Step2: Calculate moles of sucrose

Given mass of sucrose $m = 1.00\ g$. Moles of sucrose $n_{sucrose}=\frac{m}{M_{sucrose}}=\frac{1.00\ g}{342\ g/mol}\approx0.00292\ mol$.

Step3: Determine mole - ratio from balanced equation

From the balanced equation $C_{12}H_{22}O_{11}+12O_{2}
ightarrow12CO_{2}+11H_{2}O$, the mole - ratio of $C_{12}H_{22}O_{11}$ to $O_{2}$ is $1:12$.

Step4: Calculate moles of oxygen

Moles of oxygen $n_{O_{2}}=12\times n_{sucrose}=12\times0.00292\ mol = 0.03504\ mol$.

Step5: Calculate mass of oxygen

The molar mass of $O_{2}$ is $M_{O_{2}} = 32\ g/mol$. Mass of oxygen $m_{O_{2}}=n_{O_{2}}\times M_{O_{2}}=0.03504\ mol\times32\ g/mol = 1.12128\ g\approx1.12\ g$.

Answer:

$1.12$