Question

1. mole-to-mole (synthesis of water)the reaction between hydrogen gas ($\text{h}_2$) and oxygen gas ($\text{o}_2$) produces water ($\text{h}_2\text{o}$).- balanced equation: - if you completely react 4.50 moles of oxygen gas ($\text{o}_2$), how many moles of water ($\text{h}_2\text{o}$) will be produced? - moles of $\text{h}_2\text{o}$: mol2. mole-to-mass (combustion of methane)methane gas ($\text{ch}_4$) undergoes combustion, reacting with oxygen ($\text{o}_2$) to produce carbon dioxide ($\text{co}_2$) and water ($\text{h}_2\text{o}$).- balanced equation: - if 0.750 moles of methane ($\text{ch}_4$) burns completely, what mass (in grams) of carbon dioxide ($\text{co}_2$) is produced? - mass of $\text{co}_2$: g

Explanation:

Step1: Balance water synthesis equation

$$2\text{H}_2 + \text{O}_2 ightarrow 2\text{H}_2\text{O}$$

Step2: Relate $\text{O}_2$ to $\text{H}_2\text{O}$ moles

Mole ratio: $\frac{n(\text{H}_2\text{O})}{n(\text{O}_2)} = \frac{2}{1}$

Step3: Calculate $\text{H}_2\text{O}$ moles

$n(\text{H}_2\text{O}) = 4.50\ \text{mol} \times 2 = 9.00\ \text{mol}$

Step4: Balance methane combustion equation

$$\text{CH}_4 + 2\text{O}_2 ightarrow \text{CO}_2 + 2\text{H}_2\text{O}$$

Step5: Relate $\text{CH}_4$ to $\text{CO}_2$ moles

Mole ratio: $\frac{n(\text{CO}_2)}{n(\text{CH}_4)} = \frac{1}{1}$
$n(\text{CO}_2) = 0.750\ \text{mol}$

Step6: Compute molar mass of $\text{CO}_2$

$M(\text{CO}_2) = 12.01 + 2\times16.00 = 44.01\ \text{g/mol}$

Step7: Calculate mass of $\text{CO}_2$

$m(\text{CO}_2) = 0.750\ \text{mol} \times 44.01\ \text{g/mol} = 33.0\ \text{g}$

Answer:

1. Balanced Equation: $2\text{H}_2 + \text{O}_2

ightarrow 2\text{H}_2\text{O}$
Moles of $\text{H}_2\text{O}$: $9.00$ mol

1. Balanced Equation: $\text{CH}_4 + 2\text{O}_2

ightarrow \text{CO}_2 + 2\text{H}_2\text{O}$
Mass of $\text{CO}_2$: $33.0$ g