Question

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?

Answer 1

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 `CO_2`. We have:

`6H_2O + 6CO_2 -> C_6H_12O_6 + 6O_2`

We must convert the carbon dioxide to moles. We know that the molar mass of `CO_2` is `12 + 2(16) g/"mol" = 44 g/"mol"`

`88.0 g * (mol)/(44 g) = 2 mol`

Now we look at our chemical equation's mole ratios. We can see that `6` moles of `CO_2` gives `1` mole of `C_6H_12O_6` therefore, we will have `1/3` mol of glucose. The molar mass of glucose is `6(12)+ 12(1) + 6(16) = 180 g/"mol"`.

`1/3 " mol" * 180 g/"mol" = 60 g`

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

Hopefully this helps!