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

The equation is: $\text{3NiCl"_2 + "2Na"_3"PO"_4}$$\rightarrow$$\text{Ni"_3"(PO"_4)_2 + "6NaCl}$

Feb 18, 2016

You will need $\text{2.15 mol NiCl"_2}$ to produce $\text{0.715 mol Ni"_3("PO"_4")"_2}$.

Explanation:

Balanced Equation

${\text{3NiCl"_2+"2Na"_3"PO}}_{4}$$\rightarrow$$\text{Ni"_3("PO"_4")"_2+"6NaCl}$

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

Mole Ratio

$\text{3 mol NiCl"_2} :$$\text{1 mol Ni"_3("PO"_4")"_2}$

Multiply the given moles of $\text{Ni"_3("PO"_4")"_2}$ times the mole ratio with $\text{NiCl"_2}$ in the numerator.

$0.715 \cancel{\text{mol Ni"_3("PO"_4")"_2}}$$\times$(3"mol NiCl"_2)/((1cancel("mol Ni"_3("PO"_4")"_2"))$=$$\text{2.15 mol NiCl"_2}$

You will need $\text{2.15 mol NiCl"_2}$ to produce $\text{0.715 mol Ni"_3("PO"_4")"_2}$.

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

$2.15 \cancel{\text{mol NiCl"_2xx(129.5994"g NiCl"_2)/(1cancel"mol NiCl"_2)="279 g NiCl"_2}}$ rounded to three significant figures.

So in order to obtain $\text{2.15 mol NiCl"_2}$, you would use $\text{279 g NiCl"_2}$.