Answer: 49.05 g
Explanation: \(6CO_2+6H_2O\rightarrow C_6H_{12}O_6+6O_2\)
Given: Molar mass of \(C_6H_{12}O_6\)= 180.18 g/mol
Molar mass of \(CO_2\) = 44.01 g/mol
From the given balanced equation:
6 moles of carbon dioxide \((CO_2)\) produce 1 mole of glucose \((C_6H_{12}O_6)\)
\(6\times 44.01=264.06g\) of carbon dioxide\((CO_2)\) produce 180.18 of glucose \((C_6H_{12}O_6)\)
71.89 g of carbon dioxide\((CO_2)\) produce=\(\frac{180.18}{264.06}\times 71.89=49.05g\) of glucose \((C_6H_{12}O_6)\)
