Here are the step-by-step solutions for the three problems on the worksheet.
Problem 1: Electrolysis of Water
Goal: Find out how many grams of hydrogen gas ($H_2$) are produced from 36 g of water ($H_2O$).
Step 1: Write the balanced chemical equation.
Water breaks down into hydrogen and oxygen.
$$2 H_2O_{(l)} \rightarrow 2 H_2{(g)} + 1 O_2{(g)}$$
Step 2: Identify Molar Masses.
* Hydrogen ($H$) has a mass of approx $1.0$ g/mol. So, $H_2 = 2.0$ g/mol.
* Oxygen ($O$) has a mass of approx $16.0$ g/mol.
* Water ($H_2O$) = $(2 \times 1.0) + 16.0 = 18.0$ g/mol.
Step 3: Convert grams of Water to moles.
$$36 \text{ g } H_2O \times \frac{1 \text{ mol } H_2O}{18.0 \text{ g } H_2O} = 2.0 \text{ mol } H_2O$$
Step 4: Use the mole ratio to find moles of Hydrogen.
Looking at the balanced equation, 2 moles of water produce 2 moles of hydrogen (a 1:1 ratio).
$$2.0 \text{ mol } H_2O \times \frac{2 \text{ mol } H_2}{2 \text{ mol } H_2O} = 2.0 \text{ mol } H_2$$
Step 5: Convert moles of Hydrogen to grams.
$$2.0 \text{ mol } H_2 \times \frac{2.0 \text{ g } H_2}{1 \text{ mol } H_2} = 4.0 \text{ g } H_2$$
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Problem 2: Making Sodium Chloride
Goal: Find out how many grams of sodium ($Na$) are needed to make 581 g of sodium chloride ($NaCl$).
Step 1: Write the balanced chemical equation.
Sodium reacts with chlorine gas to make salt.
$$2 Na_{(s)} + 1 Cl_2{(g)} \rightarrow 2 NaCl_{(s)}$$
Step 2: Identify Molar Masses.
* Sodium ($Na$) = $23.0$ g/mol.
* Chlorine ($Cl$) = $35.5$ g/mol.
* Sodium Chloride ($NaCl$) = $23.0 + 35.5 = 58.5$ g/mol.
Step 3: Convert grams of Product ($NaCl$) to moles.
$$581 \text{ g } NaCl \times \frac{1 \text{ mol } NaCl}{58.5 \text{ g } NaCl} \approx 9.93 \text{ mol } NaCl$$
Step 4: Use the mole ratio to find moles of Sodium ($Na$).
The equation shows that 2 moles of $Na$ make 2 moles of $NaCl$ (a 1:1 ratio).
$$9.93 \text{ mol } NaCl \times \frac{2 \text{ mol } Na}{2 \text{ mol } NaCl} = 9.93 \text{ mol } Na$$
Step 5: Convert moles of Sodium to grams.
$$9.93 \text{ mol } Na \times \frac{23.0 \text{ g } Na}{1 \text{ mol } Na} = 228.39 \text{ g } Na$$
Rounding to 3 significant figures (based on the given "581 g"):
228 g.
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Problem 3: Burning Glucose
Goal: Find out how many grams of oxygen ($O_2$) are needed to react with 180.0 g of glucose ($C_6H_{12}O_6$).
Step 1: Write the balanced chemical equation.
Glucose burns with oxygen to create carbon dioxide and water.
$$1 C_6H_{12}O_{6(s)} + 6 O_{2(g)} \rightarrow 6 CO_{2(g)} + 6 H_2O_{(g)}$$
Step 2: Identify Molar Masses.
* Glucose ($C_6H_{12}O_6$):
* $C: 6 \times 12.0 = 72.0$
* $H: 12 \times 1.0 = 12.0$
* $O: 6 \times 16.0 = 96.0$
* Total = $180.0$ g/mol.
* Oxygen Gas ($O_2$): $2 \times 16.0 = 32.0$ g/mol.
Step 3: Convert grams of Glucose to moles.
$$180.0 \text{ g Glucose} \times \frac{1 \text{ mol Glucose}}{180.0 \text{ g Glucose}} = 1.00 \text{ mol Glucose}$$
Step 4: Use the mole ratio to find moles of Oxygen.
The equation shows 1 mole of glucose needs 6 moles of oxygen.
$$1.00 \text{ mol Glucose} \times \frac{6 \text{ mol } O_2}{1 \text{ mol Glucose}} = 6.00 \text{ mol } O_2$$
Step 5: Convert moles of Oxygen to grams.
$$6.00 \text{ mol } O_2 \times \frac{32.0 \text{ g } O_2}{1 \text{ mol } O_2} = 192.0 \text{ g } O_2$$
Final Answer:
1. 4.0 g $H_2$
2. 228 g Na
3. 192.0 g $O_2$
Parent Tip: Review the logic above to help your child master the concept of stoichiometry limiting reagent worksheet.
