Stoichiometry calculations #1

How many grams of water and carbon dioxide are formed after the combustion of 1.275 grams of sucrose?

SOLUTION:
Recall that the chemical formula of sucrose is: $C_{12}H_{22}O_{11}$, and its molecular weight is: $MW=342.34gr/mol$.

The equation of combustion of sucrose is as follows:
$$C_{12}H_{22}O_{11}+O_2 \rightarrow CO_2+H_2O$$

This reaction must be balanced first.
Being a redox reaction, we can write the two separate reactions of oxidation and reduction:

$$\overset{0}{C_{12}}\overset{+1}{H_{22}}\overset{-2}{O_{11}}+\overset{0}{O_2} \rightarrow \overset{+4}{C}\overset{-2}{O_2}+\overset{+1}{H_2}\overset{-2}{O}$$

$$\begin{cases} \overset{0}{O_2}+4e^- \rightarrow \overset{-2}{O_2} & \quad \mbox{(x12)}\\\\ \overset{0}{C_{12}} \rightarrow 12\overset{+4}{C}+12 \cdot 4e^- \end{cases}\quad +$$
$$\rule[0.3cm]{8cm}{0.01cm} $$
$$= \quad \overset{0}{C_{12}}+12\overset{0}{O_2} \rightarrow 12\overset{+4}{C}\overset{-2}{O_2}$$

The balanced equation is:

$$C_{12}H_{22}O_{11}+12O_2 \rightarrow 12CO_2+11H_2O$$

The number of moles of sucrose contained in 1.275 grams is:

$$n_{C_{12}H_{22}O_{11}}=\frac{1.275gr}{342.34gr/mol}=0.0037mol$$

From the stoichiometry of the reaction, we know that for every mole of sucrose that is burned, you obtain 12 moles of carbon dioxide and 11 moles of water. Then we have:

$$n_{CO_2}=12n_{C_{12}H_{22}O_{11}}=0.0447mol$$

$$n_{H_2O}=11n_{C_{12}H_{22}O_{11}}=0.0410mol$$

Knowing the molecular weight of carbon dioxide (44gr/mol) and water (18gr/mol), we can calculate the weight (in grams) that is formed for each compound:

$$m_{CO_2}=44gr/mol \cdot 0.0447mol=1.9668gr$$

$$m_{H_2O}=18gr/mol \cdot 0.0410mol=0.7380gr$$

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PROBLEM 21:
What mass of calcium carbonate would be formed if 248.6g of carbon dioxide were exhaled into limewater, $Ca(OH)_2$? How many grams of calcium would be needed to form that amount of calcium carbonate? Assume 100% yield in each reaction.