Question

6H2O + 6CO2 --> C6H12O6 + 6O2 How many grams of Glucose are made if you start with 88.0g of CO2?

Answer 1

Approximately `60 \ "g of" \ C_6H_12O_6`.

Explanation:

We have the balanced equation (without state symbols):

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

So, we would need six moles of carbon dioxide to fully produce one mole of glucose.

Here, we got `88 \ "g"` of carbon dioxide, and we need to convert it into moles.

Carbon dioxide has a molar mass of `44 \ "g/mol"`. So here, there exist

`(88color(red)cancelcolor(black)"g")/(44color(red)cancelcolor(black)"g""/mol")=2 \ "mol"`

Since there are two moles of `CO_2`, we can produce `2/6*1=1/3` moles of glucose `(C_6H_12O_6)`.

We need to find the mass of the glucose produced, so we multiply the number of moles of glucose by its molar mass.

Glucose has a molar mass of `180.156 \ "g/mol"`. So here, the mass of glucose produced is

`1/3color(red)cancelcolor(black)"mol"*(180.156 \ "g")/(color(red)cancelcolor(black)"mol")~~60 \ "g"` to the nearest whole number.

So, approximately `60` grams of glucose will be produced.