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

Mar 14, 2018

Approximately $60 \setminus \text{g of} \setminus {C}_{6} {H}_{12} {O}_{6}$.

#### Explanation:

We have the balanced equation (without state symbols):

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

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

Here, we got $88 \setminus \text{g}$ of carbon dioxide, and we need to convert it into moles.

Carbon dioxide has a molar mass of $44 \setminus \text{g/mol}$. So here, there exist

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

Since there are two moles of $C {O}_{2}$, we can produce $\frac{2}{6} \cdot 1 = \frac{1}{3}$ moles of glucose $\left({C}_{6} {H}_{12} {O}_{6}\right)$.

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 $180.156 \setminus \text{g/mol}$. So here, the mass of glucose produced is

$\frac{1}{3} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mol"*(180.156 \ "g")/(color(red)cancelcolor(black)"mol")~~60 \ "g}}}}$ to the nearest whole number.

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