# Question #2f32f

Nov 30, 2017

$C a S {O}_{4}$ is soluble in water, for the most part.

$C a C {O}_{3}$ is insoluble in water, for the most part. $C a {\left(N {O}_{3}\right)}_{2}$ is soluble in water, for the most part.

$B a S {O}_{4}$ is insoluble in water, for the most part. ${K}_{2} S {O}_{4}$ is soluble in water for the most part.

Hence, if we're after a homogeneous solution without precipitate,

$C a {\left(N {O}_{3}\right)}_{2} \left(a q\right) + {K}_{2} S {O}_{4} \left(a q\right) \to C a S {O}_{4} \left(s\right) + 2 K N {O}_{3} \left(a q\right)$

is probably the reaction.