Question

In an extremely explosive reaction between Glucose (C6H12O6) and Oxygen, 294.7g of water is produced when 500 grams of glucose is combusted. What is the percent yield of this reaction? (remember that the products of a combustion reaction are water and carbon dioxide)

Answer 1

Answer on Question #85059 – Chemistry – Other

Task:

In an extremely explosive reaction between Glucose $(C_6H_{12}O_6)$ and Oxygen, 294.7 g of water is produced when 500 grams of glucose is combusted. What is the percent yield of this reaction? (remember that the products of a combustion reaction are water and carbon dioxide).

Solution:

The equation for the combustion of glucose is:

$$C_6H_{12}O_6 + 6O_2 = 6CO_2 + 6H_2O$$

Molar Masses of Glucose and Water:

$$M(C_6H_{12}O_6) = 6*Ar(C) + 12*Ar(H) + 6*Ar(O);$$ $$M(C_6H_{12}O_6) = 6*12 + 12*1 + 6*16 = 180\ \mathrm{g/mol}$$ $$M(H_2O) = 2*Ar(H) + Ar(O) = 2*1 + 16 = 18\ \mathrm{g/mol}$$

According to the chemical reaction equation:

$$n(C_6H_{12}O_6) = \frac{n(H_2O)}{6};$$ $$\frac{m(C_6H_{12}O_6)}{M(C_6H_{12}O_6)} = \frac{m(H_2O)}{6*M(H_2O)};$$

Then,

$$m_{\text{theor}}(H_2O) = \frac{6*M(H_2O)*m(C_6H_{12}O_6)}{M(C_6H_{12}O_6)};$$ $$m_{\text{theor}}(H_2O) = \frac{6*18\ \mathrm{g/mol} * 500\ \mathrm{g}}{180\ \mathrm{g/mol}} = 300\ \mathrm{g}$$

The percentage yield is calculated by dividing the amount of the obtained desired product by the theoretical yield:

$$\text{percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} * 100\%$$ $$\begin{array}{l} m_{\text{theor}} (H_2O) = 300g \\ m_{\text{actual}} (H_2O) = 294.7g \\ \end{array}$$

Since less than what was calculated was actually produced, it means that the reaction's percent yield must be smaller than 100%.

$$\begin{array}{l} \% \text{yield} = \frac{m_{\text{actual}} (H_2O)}{m_{\text{theor}} (H_2O)} * 100\% \\ \% \text{yield} = \frac{294.7g}{300g} * 100\% = 98.23\% \\ \% \text{yield} = 98.23\% \\ \end{array}$$

Answer: 98.23% is the percent yield of this reaction.

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