Question

\in our bodies, sugar is broken down with oxygen to produce water and carbon dioxide.\ for the following reaction, 6.57 grams of glucose (c₆h₁₂o₆) are mixed with excess oxygen gas. the reaction yields 7.41 grams of carbon dioxide. glucose (c₆h₁₂o₆)(s) + oxygen (g)→carbon dioxide (g) + water (l) what is the theoretical yield of carbon dioxide? grams what is the percent yield of carbon dioxide? % the equation for this reaction is: c₆h₁₂o₆(s) + 6o₂(g)→6co₂(g) + 6h₂o(l)

Explanation:

Step1: Calculate moles of glucose

Molar mass of \(C_6H_{12}O_6=(6\times12)+(12\times1)+(6\times16)=180\space g/mol\)
Moles of \(C_6H_{12}O_6=\frac{6.57\space g}{180\space g/mol}= 0.0365\space mol\)

Step2: Calculate moles of \(CO_2\) using mole ratio

From \(C_6H_{12}O_6 + 6O_2
ightarrow6CO_2+6H_2O\), mole ratio \(C_6H_{12}O_6:CO_2 = 1:6\)
Moles of \(CO_2=0.0365\space mol\times6 = 0.219\space mol\)

Step3: Calculate theoretical yield of \(CO_2\)

Molar mass of \(CO_2 = 12+(2\times16)=44\space g/mol\)
Theoretical yield \(=0.219\space mol\times44\space g/mol = 9.636\space g\)

Step4: Calculate percent yield

Percent yield \(=\frac{actual\space yield}{theoretical\space yield}\times100=\frac{7.41\space g}{9.636\space g}\times100\approx76.9\%\)

Answer:

Theoretical yield: \(9.64\) grams (rounded to two decimal places)
Percent yield: \(76.9\%\)