Question

green plants use light from the sun to drive photosynthesis, a chemical reaction in which liquid water and carbon dioxide gas form aqueous glucose ($\ce{c_{6}h_{12}o_{6}}$) and oxygen ($\ce{o_{2}}$) gas. calculate the moles of carbon dioxide needed to produce 2.0 mol of oxygen. be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.

Explanation:

Step1: Write the photosynthesis reaction

The balanced chemical equation for photosynthesis is:
$$6\ce{CO_2} + 6\ce{H_2O} \xrightarrow{\text{light}} \ce{C_6H_{12}O_6} + 6\ce{O_2}$$

Step2: Determine the mole ratio of $\ce{CO_2}$ to $\ce{O_2}$

From the balanced equation, the mole ratio of $\ce{CO_2}$ to $\ce{O_2}$ is $6:6$ (or simplified, $1:1$? Wait, no—wait, 6 moles of $\ce{CO_2}$ produce 6 moles of $\ce{O_2}$? Wait, no: 6 $\ce{CO_2}$ → 6 $\ce{O_2}$? Wait, no, let's check again. The equation is:

$6\ce{CO_2} + 6\ce{H_2O}
ightarrow \ce{C_6H_{12}O_6} + 6\ce{O_2}$

So moles of $\ce{CO_2}$ : moles of $\ce{O_2}$ = $6:6$ = $1:1$? Wait, no—wait, 6 moles $\ce{CO_2}$ produce 6 moles $\ce{O_2}$. So the ratio is $\frac{\text{moles of } \ce{CO_2}}{\text{moles of } \ce{O_2}} = \frac{6}{6} = 1$? Wait, no, that can't be. Wait, no: 6 moles $\ce{CO_2}$ → 6 moles $\ce{O_2}$. So if we have 2.0 mol $\ce{O_2}$, how much $\ce{CO_2}$?

Wait, let's re-express the ratio. From the equation:

Moles of $\ce{CO_2}$ needed = (Moles of $\ce{O_2}$ produced) × $\frac{\text{moles of } \ce{CO_2}}{\text{moles of } \ce{O_2}}$

From the balanced equation, 6 moles $\ce{CO_2}$ produce 6 moles $\ce{O_2}$. So the ratio is $\frac{6}{6} = 1$? Wait, no—wait, 6 moles $\ce{CO_2}$ → 6 moles $\ce{O_2}$. So for every 6 moles $\ce{O_2}$, we need 6 moles $\ce{CO_2}$. So the ratio is 1:1? Wait, that seems off. Wait, no, let's check the formula of glucose: $\ce{C_6H_{12}O_6}$. So in glucose, there are 6 C, 12 H, 6 O. The oxygen in $\ce{O_2}$ comes from $\ce{H_2O}$? Wait, no, the standard photosynthesis equation: the oxygen in $\ce{O_2}$ is from water, and the oxygen in glucose is from $\ce{CO_2}$. Wait, maybe I messed up the equation. Let me confirm:

The correct balanced equation for photosynthesis is:

$6\ce{CO_2(g)} + 6\ce{H_2O(l)} \xrightarrow{\text{light}} \ce{C_6H_{12}O_6(aq)} + 6\ce{O_2(g)}$

Yes, that's correct. So 6 moles of $\ce{CO_2}$ react with 6 moles of $\ce{H_2O}$ to produce 1 mole of glucose and 6 moles of $\ce{O_2}$.

So the mole ratio of $\ce{CO_2}$ to $\ce{O_2}$ is $6:6$ = $1:1$? Wait, 6 moles $\ce{CO_2}$ produce 6 moles $\ce{O_2}$. So the ratio is $\frac{\text{moles } \ce{CO_2}}{\text{moles } \ce{O_2}} = \frac{6}{6} = 1$. So if we have 2.0 moles of $\ce{O_2}$, the moles of $\ce{CO_2}$ needed is $2.0 \, \text{mol} \, \ce{O_2} \times \frac{6 \, \text{mol} \, \ce{CO_2}}{6 \, \text{mol} \, \ce{O_2}} = 2.0 \, \text{mol} \, \ce{CO_2}$? Wait, that can't be right. Wait, no—wait, 6 moles $\ce{CO_2}$ produce 6 moles $\ce{O_2}$, so the ratio is 1:1. So 2.0 mol $\ce{O_2}$ would require 2.0 mol $\ce{CO_2}$? But that seems too simple. Wait, maybe I made a mistake. Wait, let's think about the number of O atoms. In $\ce{O_2}$, each mole has 2 O atoms. In $\ce{CO_2}$, each mole has 2 O atoms. Wait, no, the equation is balanced, so the mole ratio is 6:6. So yes, 6 moles $\ce{CO_2}$ → 6 moles $\ce{O_2}$. So the ratio is 1:1. Therefore, moles of $\ce{CO_2}$ = moles of $\ce{O_2}$ × $\frac{6}{6}$ = moles of $\ce{O_2}$ × 1.

Wait, but let's check with the glucose formula. Glucose has 6 O atoms. The $\ce{O_2}$ has 6 O atoms (6 moles $\ce{O_2}$ is 12 moles O atoms). The $\ce{CO_2}$ has 6 moles × 2 = 12 O atoms, and $\ce{H_2O}$ has 6 moles × 1 = 6 O atoms. Wait, no, glucose has 6 O atoms. So total O from $\ce{CO_2}$: 6 moles $\ce{CO_2}$ × 2 O = 12 O. From $\ce{H_2O}$: 6 moles × 1 O = 6 O. Total O: 18. Glucose has 6 O, $\ce{O_2}$ has 6 moles × 2 O = 12 O. 6 + 12 = 18. So that balances. So the equation is correct.

Therefore, the mole ratio of $\ce{CO_2}$ to $\ce{O_2}$ is 6:6, or 1:1. Wait, no—6 moles $\ce{CO_2}$ produce 6 moles $\ce{O_2}$. So the ratio is 6/6 = 1. So if we have 2.0 mol $\ce{O_2}$, we need 2.0 mol $\ce{CO_2}$? But that seems too straightforward. Wait, maybe I messed up the ratio. Wait, let's do it again.

From the balanced equation:

$6\ce{CO_2} + 6\ce{H_2O}
ightarrow \ce{C_6H_{12}O_6} + 6\ce{O_2}$

So, moles of $\ce{CO_2}$ : moles of $\ce{O_2}$ = 6 : 6 = 1 : 1.

Therefore, moles of $\ce{CO_2}$ = moles of $\ce{O_2}$ × (6/6) = 2.0 mol × 1 = 2.0 mol? Wait, but the problem says "round it to 2 significant digits". 2.0 has two significant digits, so 2.0 mol? Wait, but maybe I made a mistake. Wait, no—wait, 6 moles $\ce{CO_2}$ produce 6 moles $\ce{O_2}$. So the ratio is 1:1. So if you have 2.0 mol $\ce{O_2}$, you need 2.0 mol $\ce{CO_2}$.

Wait, but let's check with another approach. Suppose we have 2.0 mol $\ce{O_2}$. The mole ratio of $\ce{CO_2}$ to $\ce{O_2}$ is 6/6 = 1. So moles $\ce{CO_2}$ = 2.0 mol $\ce{O_2}$ × (6 mol $\ce{CO_2}$ / 6 mol $\ce{O_2}$) = 2.0 mol.

Yes, that makes sense. So the calculation is straightforward.

Step3: Round to 2 significant digits

The given moles of $\ce{O_2}$ is 2.0 mol (two significant digits), so the moles of $\ce{CO_2}$ is also 2.0 mol (since the ratio is 1:1 and we round to two significant digits).

Answer:

$2.0\ \text{mol}$