Question

How many moles of `"NiCl"_2` are required to produce `"0.715 moles Ni"_3"(PO"_4)_2"`?

The equation is:

`"3NiCl"_2 + "2Na"_3"PO"_4"` `rarr` `"Ni"_3"(PO"_4)_2 + "6NaCl"`

Answer 1

You will need `"2.15 mol NiCl"_2"` to produce `"0.715 mol Ni"_3("PO"_4")"_2"`.

Explanation:

Balanced Equation

`"3NiCl"_2+"2Na"_3"PO"_4` `rarr` `"Ni"_3("PO"_4")"_2+"6NaCl"`

Since we are starting with moles of `"Ni"_3("PO"_4")` and ending with moles of `"NiCl"_2"`, we need the mole ratio between these compounds from the balanced equation.

Mole Ratio

`"3 mol NiCl"_2":` `"1 mol Ni"_3("PO"_4")"_2"`

Multiply the given moles of `"Ni"_3("PO"_4")"_2"` times the mole ratio with `"NiCl"_2"` in the numerator.

`0.715cancel("mol Ni"_3("PO"_4")"_2")` `xx` `(3"mol NiCl"_2)/((1cancel("mol Ni"_3("PO"_4")"_2"))` `=` `"2.15 mol NiCl"_2"`

You will need `"2.15 mol NiCl"_2"` to produce `"0.715 mol Ni"_3("PO"_4")"_2"`.

Since it's easier to work with mass, you can convert moles `"NiCl"_2"` to mass in grams by multiplying the calculated moles `"NiCl"_2"` times its molar mass, `"129.5994 g/mol"`. https://pubchem.ncbi.nlm.nih.gov/compound/24385

`2.15cancel"mol NiCl"_2xx(129.5994"g NiCl"_2)/(1cancel"mol NiCl"_2)="279 g NiCl"_2"` rounded to three significant figures.

So in order to obtain `"2.15 mol NiCl"_2"`, you would use `"279 g NiCl"_2"`.