Solution-Stoichiometry Worksheet Key PDF | PDF | Sodium Hydroxide ... - Free Printable

Here are the step-by-step solutions for the problems on the worksheet.

Problem 1

Question: How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potassium chromate?

Step 1: Write the balanced chemical equation.
$$2 \text{AgNO}_3(aq) + \text{K}_2\text{CrO}_4(aq) \rightarrow \text{Ag}_2\text{CrO}_4(s) + 2 \text{KNO}_3(aq)$$

Step 2: Calculate moles of each reactant.
* Silver Nitrate ($\text{AgNO}_3$):
$$0.150 \text{ L} \times 0.500 \text{ mol/L} = 0.075 \text{ moles AgNO}_3$$
* Potassium Chromate ($\text{K}_2\text{CrO}_4$):
$$0.100 \text{ L} \times 0.400 \text{ mol/L} = 0.040 \text{ moles K}_2\text{CrO}_4$$

Step 3: Determine the limiting reactant.
We calculate how much product ($\text{Ag}_2\text{CrO}_4$) each reactant could make.
* From $\text{AgNO}_3$: The ratio is 2:1.
$$0.075 \text{ mol AgNO}_3 \times \frac{1 \text{ mol Ag}_2\text{CrO}_4}{2 \text{ mol AgNO}_3} = 0.0375 \text{ mol Ag}_2\text{CrO}_4$$
* From $\text{K}_2\text{CrO}_4$: The ratio is 1:1.
$$0.040 \text{ mol K}_2\text{CrO}_4 \times \frac{1 \text{ mol Ag}_2\text{CrO}_4}{1 \text{ mol K}_2\text{CrO}_4} = 0.040 \text{ mol Ag}_2\text{CrO}_4$$

Since $0.0375$ is smaller than $0.040$, Silver Nitrate is the limiting reactant. We use $0.0375$ moles for the final calculation.

Step 4: Convert moles of product to grams.
Molar Mass of $\text{Ag}_2\text{CrO}_4 \approx 331.74 \text{ g/mol}$.
$$0.0375 \text{ mol} \times 331.74 \text{ g/mol} = 12.44 \text{ g}$$

Rounding to 3 significant figures (based on the given values): 12.4 g

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Problem 2

Question: How many mL of 0.280 M barium nitrate are required to precipitate as barium sulfate all the sulfate ions from 25.0 mL of 0.350 M aluminum sulfate?

Step 1: Write the balanced chemical equation.
$$3 \text{Ba(NO}_3)_2(aq) + \text{Al}_2(\text{SO}_4)_3(aq) \rightarrow 3 \text{BaSO}_4(s) + 2 \text{Al(NO}_3)_3(aq)$$

Step 2: Calculate moles of Aluminum Sulfate.
$$0.0250 \text{ L} \times 0.350 \text{ mol/L} = 0.00875 \text{ moles Al}_2(\text{SO}_4)_3$$

Step 3: Calculate moles of Barium Nitrate needed.
The mole ratio is 3 moles of Barium Nitrate for every 1 mole of Aluminum Sulfate.
$$0.00875 \text{ mol Al}_2(\text{SO}_4)_3 \times \frac{3 \text{ mol Ba(NO}_3)_2}{1 \text{ mol Al}_2(\text{SO}_4)_3} = 0.02625 \text{ moles Ba(NO}_3)_2$$

Step 4: Calculate volume of Barium Nitrate solution.
$$\text{Volume} = \frac{\text{moles}}{\text{Molarity}} = \frac{0.02625 \text{ mol}}{0.280 \text{ mol/L}} = 0.09375 \text{ L}$$

Convert to mL:
$$0.09375 \text{ L} \times 1000 \text{ mL/L} = 93.75 \text{ mL}$$

Rounding to 3 significant figures: 93.8 mL

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Problem 3

Question: 25.0 mL of 0.350 M NaOH are added to 45.0 mL of 0.125 M copper (II) sulfate. How many grams of copper (II) hydroxide will precipitate?

Step 1: Write the balanced chemical equation.
$$2 \text{NaOH}(aq) + \text{CuSO}_4(aq) \rightarrow \text{Cu(OH)}_2(s) + \text{Na}_2\text{SO}_4(aq)$$

Step 2: Calculate moles of each reactant.
* NaOH:
$$0.0250 \text{ L} \times 0.350 \text{ mol/L} = 0.00875 \text{ moles NaOH}$$
* $\text{CuSO}_4$:
$$0.0450 \text{ L} \times 0.125 \text{ mol/L} = 0.005625 \text{ moles CuSO}_4$$

Step 3: Determine the limiting reactant.
* From NaOH (Ratio 2:1):
$$0.00875 \text{ mol NaOH} \times \frac{1 \text{ mol Cu(OH)}_2}{2 \text{ mol NaOH}} = 0.004375 \text{ mol Cu(OH)}_2$$
* From $\text{CuSO}_4$ (Ratio 1:1):
$$0.005625 \text{ mol CuSO}_4 \times \frac{1 \text{ mol Cu(OH)}_2}{1 \text{ mol CuSO}_4} = 0.005625 \text{ mol Cu(OH)}_2$$

NaOH produces less product, so it is the limiting reactant. We use $0.004375$ moles.

Step 4: Convert moles of product to grams.
Molar Mass of $\text{Cu(OH)}_2 \approx 97.57 \text{ g/mol}$.
$$0.004375 \text{ mol} \times 97.57 \text{ g/mol} = 0.4268... \text{ g}$$

Rounding to 3 significant figures: 0.427 g

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Problem 4

Question: What volume of 0.415 M silver nitrate will be required to precipitate as silver bromide all the bromide ion in 35.0 mL of 0.128 M calcium bromide?

Step 1: Write the balanced chemical equation.
$$2 \text{AgNO}_3(aq) + \text{CaBr}_2(aq) \rightarrow \text{Ca(NO}_3)_2(aq) + 2 \text{AgBr}(s)$$

Step 2: Calculate moles of Calcium Bromide ($\text{CaBr}_2$).
$$0.0350 \text{ L} \times 0.128 \text{ mol/L} = 0.00448 \text{ moles CaBr}_2$$

Step 3: Calculate moles of Silver Nitrate ($\text{AgNO}_3$) needed.
The ratio is 2 moles of $\text{AgNO}_3$ for every 1 mole of $\text{CaBr}_2$.
$$0.00448 \text{ mol CaBr}_2 \times \frac{2 \text{ mol AgNO}_3}{1 \text{ mol CaBr}_2} = 0.00896 \text{ moles AgNO}_3$$

Step 4: Calculate volume of Silver Nitrate solution.
$$\text{Volume} = \frac{\text{moles}}{\text{Molarity}} = \frac{0.00896 \text{ mol}}{0.415 \text{ mol/L}} \approx 0.02159 \text{ L}$$

Convert to Liters (standard scientific notation usually keeps 3 sig figs):
0.0216 L (or 21.6 mL)

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Problem 5

Question: What volume of 0.496 M HCl is required to neutralize 20.0 mL of 0.809 M sodium hydroxide?

Step 1: Write the balanced chemical equation.
$$\text{HCl}(aq) + \text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)$$

Step 2: Calculate moles of Sodium Hydroxide (NaOH).
$$0.0200 \text{ L} \times 0.809 \text{ mol/L} = 0.01618 \text{ moles NaOH}$$

Step 3: Calculate moles of HCl needed.
The ratio is 1:1.
$$0.01618 \text{ moles NaOH} = 0.01618 \text{ moles HCl}$$

Step 4: Calculate volume of HCl solution.
$$\text{Volume} = \frac{\text{moles}}{\text{Molarity}} = \frac{0.01618 \text{ mol}}{0.496 \text{ mol/L}} \approx 0.03262 \text{ L}$$

Rounding to 3 significant figures: 0.0326 L (or 32.6 mL)

Final Answer:
1. 12.4 g Ag₂CrO₄
2. 93.8 mL Ba(NO₃)₂
3. 0.427 g Cu(OH)₂
4. 0.0216 L AgNO₃
5. 0.0326 L HCl