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"#

1 Answer
Feb 18, 2016

Answer:

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"#.