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

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} C_6H_(12)O_6 =(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=
`(6)/(1)* 0.17=1.0mole` of
`CO_2`

Mass of
`CO_2=moles* {\text {Molar mass}}=1.0moles* 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.