Question

part 3 of 4

1. the white precipitate formed in the reaction between $(nh_4)_3po_4$ and $cdbr_2$ is $square$.
2. the white precipitate formed in the reaction between $lioh$ and $mncl_2$ is $square$.

Explanation:

Step1: Identify first reaction products

Reactants: $\ce{(NH4)3PO4}$ and $\ce{CdBr2}$. Swap ions: $\ce{Cd3(PO4)2}$ and $\ce{NH4Br}$. $\ce{NH4Br}$ is soluble; $\ce{Cd3(PO4)2}$ is insoluble white precipitate.

Step2: Identify second reaction products

Reactants: $\ce{LiOH}$ and $\ce{MnCl2}$. Swap ions: $\ce{Mn(OH)2}$ and $\ce{LiCl}$. $\ce{LiCl}$ is soluble; $\ce{Mn(OH)2}$ is insoluble white precipitate.

Answer:

1. The white precipitate formed in the reaction between $\ce{(NH4)3PO4}$ and $\ce{CdBr2}$ is $\ce{Cd3(PO4)2}$.
2. The white precipitate formed in the reaction between $\ce{LiOH}$ and $\ce{MnCl2}$ is $\ce{Mn(OH)2}$.