Question

Manganese (V) carbonate + nickel (II) phosphite -->

Answer 1

`3Mn_2(CO_3)_5+5Ni_3(PO_3)_2\rightarrow 15NiCO_3+2Mn_3(PO_3)_5`

Step-by-step explanation:

Hello there!

In this case, for the given reactants side in this chemical reaction, it is possible for us to firstly write the left side of the undergoing chemical equation as shown below:

`Mn_2(CO_3)_5+Ni_3(PO_3)_2\rightarrow NiCO_3+Mn_3(PO_3)_5`

Which means that the products are nickel (II) carbonate and manganese (V) phosphite. Next, we balance the reaction as shown below:

`3Mn_2(CO_3)_5+5Ni_3(PO_3)_2\rightarrow 15NiCO_3+2Mn_3(PO_3)_5`

Which makes 15 carbonate ions, 10 phosphite ions, 15 nicked (II) ions and 6 manganese (V) on both sides of the chemical equation.

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