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: $$6\ce{CO_2} + 6\ce{H_2O} \xrightarrow{\text{light}} \ce{C_6H_{12}O_6} + 6\ce{O_2}$$
From the equation, the mole ratio of $\ce{CO_2}$ to $\ce{O_2}$ is $6:6 = 1:1$? Wait, no, wait. Wait, 6 moles of $\ce{CO_2}$ produce 6 moles of $\ce{O_2}$. So the mole ratio of $\ce{CO_2}$ to $\ce{O_2}$ is $6/6 = 1$? Wait, no, let's check again. The equation is 6 $\ce{CO_2}$ + 6 $\ce{H_2O}$ makes 1 $\ce{C_6H_{12}O_6}$ and 6 $\ce{O_2}$. So moles of $\ce{CO_2}$ : moles of $\ce{O_2}$ = 6 : 6 = 1:1? Wait, no, 6 moles of $\ce{CO_2}$ produce 6 moles of $\ce{O_2}$. So the ratio of $\ce{CO_2}$ to $\ce{O_2}$ is 6/6 = 1. Wait, but let's see: if we have 2.0 mol of $\ce{O_2}$, how much $\ce{CO_2}$?

Wait, the stoichiometry: for every 6 moles of $\ce{O_2}$ produced, 6 moles of $\ce{CO_2}$ are consumed. So the mole ratio of $\ce{CO_2}$ to $\ce{O_2}$ is 6/6 = 1. So moles of $\ce{CO_2}$ = moles of $\ce{O_2}$ × (6 moles $\ce{CO_2}$ / 6 moles $\ce{O_2}$) = 2.0 mol $\ce{O_2}$ × (6/6) = 2.0 mol? Wait, that can't be right. Wait, no, wait the equation: 6 $\ce{CO_2}$ + 6 $\ce{H_2O}$ → $\ce{C_6H_{12}O_6}$ + 6 $\ce{O_2}$. So 6 moles $\ce{CO_2}$ produce 6 moles $\ce{O_2}$. So the ratio of $\ce{CO_2}$ to $\ce{O_2}$ is 6/6 = 1. So if we have 2.0 mol $\ce{O_2}$, then moles of $\ce{CO_2}$ needed is 2.0 mol × (6/6) = 2.0 mol? Wait, but maybe I made a mistake. Wait, let's check the coefficients. 6 $\ce{CO_2}$ (g) + 6 $\ce{H_2O}$ (l) → $\ce{C_6H_{12}O_6}$ (aq) + 6 $\ce{O_2}$ (g). So moles of $\ce{CO_2}$ : moles of $\ce{O_2}$ = 6 : 6 = 1:1. So yes, 2.0 mol $\ce{O_2}$ would require 2.0 mol $\ce{CO_2}$? Wait, but that seems too simple. Wait, maybe the ratio is 6 moles $\ce{CO_2}$ for 6 moles $\ce{O_2}$, so 1:1. So if we have 2.0 mol $\ce{O_2}$, then moles of $\ce{CO_2}$ is 2.0 mol × (6 mol $\ce{CO_2}$ / 6 mol $\ce{O_2}$) = 2.0 mol. But let's confirm.

Wait, let's do the stoichiometry properly. The balanced equation is:

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

So the mole ratio of $\ce{CO_2}$ to $\ce{O_2}$ is $\frac{n_{\ce{CO_2}}}{n_{\ce{O_2}}} = \frac{6}{6} = 1$.

Given $n_{\ce{O_2}} = 2.0$ mol, we can solve for $n_{\ce{CO_2}}$:

$$n_{\ce{CO_2}} = n_{\ce{O_2}} \times \frac{6\ \text{mol}\ \ce{CO_2}}{6\ \text{mol}\ \ce{O_2}}$$

Substitute $n_{\ce{O_2}} = 2.0$ mol:

$$n_{\ce{CO_2}} = 2.0\ \text{mol} \times \frac{6}{6} = 2.0\ \text{mol}$$

Wait, but that seems like the ratio is 1:1. But let's check with another approach. Suppose we have 1 mole of $\ce{O_2}$, then we need 1 mole of $\ce{CO_2}$? Wait, no, 6 moles $\ce{CO_2}$ give 6 moles $\ce{O_2}$, so 1 mole $\ce{CO_2}$ gives 1 mole $\ce{O_2}$. So yes, the ratio is 1:1. Therefore, to produce 2.0 mol of $\ce{O_2}$, we need 2.0 mol of $\ce{CO_2}$. But wait, the problem says "round it to 2 significant digits". 2.0 has two significant digits, so the answer is 2.0 mol? Wait, but maybe I messed up the mole ratio. Wait, let's re-express the equation:

6 $\ce{CO_2}$ + 6 $\ce{H_2O}$ → 1 $\ce{C_6H_{12}O_6}$ + 6 $\ce{O_2}$

So moles of $\ce{CO_2}$ = (6/6) × moles of $\ce{O_2}$ = 1 × 2.0 mol = 2.0 mol. So the moles of $\ce{CO_2}$ needed is 2.0 mol.

Step2: Check significant digits

The given amount is 2.0 mol (two significant digits), and the ratio is exact (from the balanced equation), so the answer should have two significant digits. 2.0 mol is already two significant digits. Wait, but maybe I made a mistake in the mole ratio. Wait, let's check again. The equation: 6 $\ce{CO_2}$ produces 6 $\ce{O_2}$, so the ratio is 6/6 = 1. So yes, 2.0 mol $\ce{O_2}$ requires 2.0 mol $\ce{CO_2}$.

Answer:

$\boxed{2.0\ \text{mol}}$