Question

what volume, in ml, of 0.35 m ca(no₃)₂ is required to completely react with 50.0 ml of 0.75 m k₂so₄?
ca(no₃)₂ + k₂so₄ → caso₄ + 2kno₃
? ml ca(no₃)₂
note the volume units requested.

Explanation:

Step1: Find moles of \( K_2SO_4 \)

Molarity \( M = \frac{n}{V} \), so \( n = M \times V \). Volume of \( K_2SO_4 \) is \( 50.0 \, \text{mL} = 0.0500 \, \text{L} \), molarity is \( 0.75 \, \text{M} \).
\( n_{K_2SO_4} = 0.75 \, \text{mol/L} \times 0.0500 \, \text{L} = 0.0375 \, \text{mol} \).

Step2: Relate moles of \( Ca(NO_3)_2 \) and \( K_2SO_4 \)

From the balanced equation, the mole ratio of \( Ca(NO_3)_2 \) to \( K_2SO_4 \) is \( 1:1 \). So \( n_{Ca(NO_3)_2} = n_{K_2SO_4} = 0.0375 \, \text{mol} \).

Step3: Find volume of \( Ca(NO_3)_2 \)

Using \( M = \frac{n}{V} \), rearranged \( V = \frac{n}{M} \). Molarity of \( Ca(NO_3)_2 \) is \( 0.35 \, \text{M} \), moles \( n = 0.0375 \, \text{mol} \).
\( V_{Ca(NO_3)_2} = \frac{0.0375 \, \text{mol}}{0.35 \, \text{mol/L}} \approx 0.10714 \, \text{L} \). Convert to mL: \( 0.10714 \, \text{L} \times 1000 \, \text{mL/L} \approx 107 \, \text{mL} \).

Answer:

\( 107 \) (or more precisely, using exact calculation: \( \frac{0.75 \times 50.0}{0.35} \approx 107.14 \), rounded to appropriate sig figs, 107 mL)