Question

question 19 of 25
what is the net ionic equation for the reaction between na₂so₄ and cacl₂?
a. 2na⁺(aq) + so₄²⁻(aq) + ca²⁺(aq) + 2cl⁻(aq) → caso₄(s) + 2na⁺ + 2cl⁻(aq)
b. so₄²⁻(aq) + ca²⁺(aq) → caso₄(s)
c. na₂so₄(aq) + cacl₂(aq) → caso₄(s) + 2nacl(aq)
d. 2na⁺(aq) + 2cl⁻(aq) → 2nacl(aq)

Explanation:

Step1: Write the balanced molecular equation

$Na_2SO_4(aq)+CaCl_2(aq)
ightarrow CaSO_4(s) + 2NaCl(aq)$

Step2: Write the complete ionic equation

$2Na^+(aq)+SO_4^{2 -}(aq)+Ca^{2 +}(aq)+2Cl^-(aq)
ightarrow CaSO_4(s)+2Na^+(aq)+2Cl^-(aq)$

Step3: Identify and cancel out the spectator ions

Spectator ions are $Na^+$ and $Cl^-$ which appear on both sides. After canceling them, we get the net - ionic equation.
$SO_4^{2 -}(aq)+Ca^{2 +}(aq)
ightarrow CaSO_4(s)$

Answer:

B. $SO_4^{2 -}(aq)+Ca^{2 +}(aq)
ightarrow CaSO_4(s)$