Question

What would be the theoretical yield in grams of carbon dioxide in the reaction shown below if 30 g of C6H12O6 were reacted with an excess of oxygen? C6H12O6 + O2 -> CO2 + H2O

Answer 1

Answer: Thus 44 g of $CO_2$ will be produced  if 30 g of $C_6H_{12}O_6$ were reacted with an excess of oxygen

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} C_6H_{12}O_6 =\frac{30g}{180g/mol}=0.17moles$

The balanced chemical equation is:  

$C_6H_{12}O_6 +6O_2(g)\rightarrow 6CO_2+6H_2O(g)$

As $C_6H_{12}O_6$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess reagent.

According to stoichiometry :

1 mole of $C_6H_{12}O_6$ produce =  6 moles of $CO_2$

Thus 0.17 moles of $C_6H_{12}O_6$ will produce=$\frac{6}{1}\times 0.17=1.0mole$  of $CO_2$

 Mass of $CO_2=moles\times {\text {Molar mass}}=1.0moles\times 44g/mol=44g$

Thus 44 g of $CO_2$ will be produced  if 30 g of $C_6H_{12}O_6$ were reacted with an excess of oxygen.