To determine which equation is balanced, we must ensure that the number of each type of atom on the left side (reactants) equals the number of each type of atom on the right side (products).
Let's examine each equation:
### Option A
[tex]\[
Na_2SO_4(aq) + CaCl_2(aq) \rightarrow CaSO_4(s) + 2 NaCl(aq)
\][/tex]
Left side:
- \( Na: 2 \)
- \( S: 1 \)
- \( O: 4 \)
- \( Ca: 1 \)
- \( Cl: 2 \)
Right side:
- \( Na: 2 \) (from \(2 \times NaCl\))
- \( S: 1 \)
- \( O: 4 \)
- \( Ca: 1 \)
- \( Cl: 2 \) (from \(2 \times NaCl\))
Conclusion: All atoms are balanced on both sides.
### Option B
[tex]\[
Na_2SO_4(aq) + 2 CaCl_2(aq) \rightarrow CaSO_4(s) + NaCl(aq)
\][/tex]
Left side:
- \( Na: 2 \)
- \( S: 1 \)
- \( O: 4 \)
- \( Ca: 2 \)
- \( Cl: 4 \) (from \(2 \times CaCl_2\))
Right side:
- \( Na: 1 \)
- \( S: 1 \)
- \( O: 4 \)
- \( Ca: 1 \)
- \( Cl: 1 \)
Conclusion: The counts of \( Na \), \( Ca \), and \( Cl \) are not balanced.
### Option C
[tex]\[
Na_2SO_4(aq) + CaCl_2(aq) \rightarrow CaSO_4(s) + NaCl(aq)
\][/tex]
Left side:
- \( Na: 2 \)
- \( S: 1 \)
- \( O: 4 \)
- \( Ca: 1 \)
- \( Cl: 2 \)
Right side:
- \( Na: 1 \)
- \( S: 1 \)
- \( O: 4 \)
- \( Ca: 1 \)
- \( Cl: 1 \)
Conclusion: The counts of \( Na \) and \( Cl \) are not balanced.
### Option D
[tex]\[
2 Na_2SO_4(aq) + CaCl_2(aq) \rightarrow CaSO_4(s) + 2 NaCl(aq)
\][/tex]
Left side:
- \( Na: 4 \) (from \(2 \times Na_2SO_4\))
- \( S: 2 \)
- \( O: 8 \)
- \( Ca: 1 \)
- \( Cl: 2 \)
Right side:
- \( Na: 2 \) (from \(2 \times NaCl\))
- \( S: 1 \)
- \( O: 4 \)
- \( Ca: 1 \)
- \( Cl: 2 \)
Conclusion: The counts of \( Na \), \( S \), and \( O \) are not balanced.
### Summary
From the detailed analysis:
- Option A is balanced.
- Options B, C, and D are not balanced.
Therefore, the balanced equation is:
A. [tex]\( Na_2SO_4(aq) + CaCl_2(aq) \rightarrow CaSO_4(s) + 2 NaCl(aq) \)[/tex]
