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