# With the Equation 6H2O + 6CO2 --> C6H12O6 + 6O2 How many grams of water are needed to make the above amount of glucose if you start with 88.0g of CO2?

Mar 8, 2018

The question doesn't make much sense as it is so I will show you how to find the amount of glucose, given $88.0$ g of $C {O}_{2}$. We have:

$6 {H}_{2} O + 6 C {O}_{2} \to {C}_{6} {H}_{12} {O}_{6} + 6 {O}_{2}$

We must convert the carbon dioxide to moles. We know that the molar mass of $C {O}_{2}$ is $12 + 2 \left(16\right) \frac{g}{\text{mol" = 44 g/"mol}}$

$88.0 g \cdot \frac{m o l}{44 g} = 2 m o l$

Now we look at our chemical equation's mole ratios. We can see that $6$ moles of $C {O}_{2}$ gives $1$ mole of ${C}_{6} {H}_{12} {O}_{6}$ therefore, we will have $\frac{1}{3}$ mol of glucose. The molar mass of glucose is $6 \left(12\right) + 12 \left(1\right) + 6 \left(16\right) = 180 \frac{g}{\text{mol}}$.

$\frac{1}{3} \text{ mol" * 180 g/"mol} = 60 g$

Therefore, $60$ grams of glucose will be produced.

Hopefully this helps!