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 formula: \( M = \frac{n}{V} \) (where \( M \) is molarity, \( n \) is moles, \( V \) is volume in L).
For \( K_2SO_4 \): \( M = 0.75 \, M \), \( V = 50.0 \, mL = 0.0500 \, L \).
\( n = M \times V = 0.75 \, \frac{mol}{L} \times 0.0500 \, L = 0.0375 \, mol \).

Step2: Molar ratio from reaction

Reaction: \( \text{Ca(NO}_3\text{)}_2 + \text{K}_2\text{SO}_4
ightarrow \text{CaSO}_4 + 2\text{KNO}_3 \).
Molar ratio \( \text{Ca(NO}_3\text{)}_2 : \text{K}_2\text{SO}_4 = 1:1 \).
Thus, moles of \( \text{Ca(NO}_3\text{)}_2 = \) moles of \( \text{K}_2\text{SO}_4 = 0.0375 \, mol \).

Step3: Volume of \( \text{Ca(NO}_3\text{)}_2 \)

Molarity of \( \text{Ca(NO}_3\text{)}_2 = 0.35 \, M \).
\( V = \frac{n}{M} = \frac{0.0375 \, mol}{0.35 \, \frac{mol}{L}} \approx 0.10714 \, L \).
Convert to mL: \( 0.10714 \, L \times 1000 \, \frac{mL}{L} \approx 107 \, mL \).

Answer:

\( \boxed{107} \)