Question

1. quiz limiting reactant

25 grams of silver (i) phosphide is added to 18 grams of chromium (iii) hydroxide.

1. write the complete balanced equation below: (1 point) also use this space to show your work!!

● 2ag₃p + 3cr(oh)₂ → 6agoh + 1cr₃p₂

1. what is the molar mass of each chemical in this reaction? (4 points)
2. what type of reaction is this? (1 point) ______
3. what is the limiting reactant in this reaction? (1 point) ______
4. what is the excess reactant in this reaction? (1 point) ______
5. how much of each product will form? (2 points)

Explanation:

Step1: Confirm balanced equation

The balanced reaction is:

$$2\text{Ag}_3\text{P} + 3\text{Cr(OH)}_2 ightarrow 6\text{AgOH} + \text{Cr}_3\text{P}_2$$

Step2: Calculate moles of limiting reactant

Molar mass of $\text{Cr(OH)}_2$ = $86.012\ \text{g/mol}$. Moles of $\text{Cr(OH)}_2$:
$$n_{\text{Cr(OH)}_2} = \frac{26\ \text{g}}{86.012\ \text{g/mol}} \approx 0.3023\ \text{mol}$$

Step3: Find moles of $\text{AgOH}$

Mole ratio $\text{Cr(OH)}_2:\text{AgOH} = 3:6 = 1:2$.
$$n_{\text{AgOH}} = 0.3023\ \text{mol} \times 2 = 0.6046\ \text{mol}$$

Step4: Calculate mass of $\text{AgOH}$

Molar mass of $\text{AgOH}$ = $124.876\ \text{g/mol}$.
$$m_{\text{AgOH}} = 0.6046\ \text{mol} \times 124.876\ \text{g/mol} \approx 75.5\ \text{g}$$

Step5: Find moles of $\text{Cr}_3\text{P}_2$

Mole ratio $\text{Cr(OH)}_2:\text{Cr}_3\text{P}_2 = 3:1$.
$$n_{\text{Cr}_3\text{P}_2} = \frac{0.3023\ \text{mol}}{3} \approx 0.1008\ \text{mol}$$

Step6: Calculate mass of $\text{Cr}_3\text{P}_2$

Molar mass of $\text{Cr}_3\text{P}_2$ = $217.936\ \text{g/mol}$.
$$m_{\text{Cr}_3\text{P}_2} = 0.1008\ \text{mol} \times 217.936\ \text{g/mol} \approx 21.97\ \text{g}$$

Answer:

Mass of $\text{AgOH}$: $\boldsymbol{75.5\ \text{grams}}$
Mass of $\text{Cr}_3\text{P}_2$: $\boldsymbol{22.0\ \text{grams}}$ (rounded to 3 significant figures)