Math Snap

STEP 1

1. The chemical reaction is balanced as given: $3 \mathrm{KOH} + \mathrm{H}_{3} \mathrm{PO}_{4} \rightarrow \mathrm{K}_{3} \mathrm{PO}_{4} + 3 \mathrm{H}_{2} \mathrm{O}$.
2. We are given 30.3 g of potassium hydroxide (KOH).
3. We need to find the mass of potassium phosphate ($\mathrm{K}_{3} \mathrm{PO}_{4}$) produced.
4. Molar masses:
- KOH = 39.10 (K) + 16.00 (O) + 1.01 (H) = 56.11 g/mol
- $\mathrm{K}_{3} \mathrm{PO}_{4}$ = 3(39.10) (K) + 30.97 (P) + 4(16.00) (O) = 212.27 g/mol

STEP 2

1. Calculate the moles of KOH.
2. Use stoichiometry to find the moles of $\mathrm{K}_{3} \mathrm{PO}_{4}$.
3. Calculate the mass of $\mathrm{K}_{3} \mathrm{PO}_{4}$.

STEP 3

Calculate the moles of KOH.
Using the formula:
$$\text{Moles of KOH} = \frac{\text{Mass of KOH}}{\text{Molar mass of KOH}}$$ $$\text{Moles of KOH} = \frac{30.3 \text{ g}}{56.11 \text{ g/mol}} \approx 0.540 \text{ moles}$$

STEP 4

Use stoichiometry to find the moles of $\mathrm{K}_{3} \mathrm{PO}_{4}$.
From the balanced equation, 3 moles of KOH produce 1 mole of $\mathrm{K}_{3} \mathrm{PO}_{4}$.
$$\text{Moles of } \mathrm{K}_{3} \mathrm{PO}_{4} = \frac{0.540 \text{ moles KOH}}{3} \approx 0.180 \text{ moles}$$

Calculate the mass of $\mathrm{K}_{3} \mathrm{PO}_{4}$.
Using the formula:
$$\text{Mass of } \mathrm{K}_{3} \mathrm{PO}_{4} = \text{Moles of } \mathrm{K}_{3} \mathrm{PO}_{4} \times \text{Molar mass of } \mathrm{K}_{3} \mathrm{PO}_{4}$$ $$\text{Mass of } \mathrm{K}_{3} \mathrm{PO}_{4} = 0.180 \text{ moles} \times 212.27 \text{ g/mol} \approx 38.21 \text{ g}$$ The mass of potassium phosphate produced is:
$\boxed{38.21 \text{ g}}$