Given: Mass of #"Mg"("NO"_3)_2#; chemical equation (understood)
Find: Moles of #"Ca"_3("PO"_4)_2#
Strategy:
The central part of any stoichiometry problem is to convert moles of something to moles of something else.
(a) We start with the balanced chemical equation for the reaction.
(b) We can use the molar mass of #"Mg"("NO"_3)_2# to find the moles of #"Mg"("NO"_3)_2#.
(c) We then use the molar ratio from the equation to convert moles of #"Mg"("NO"_3)_2# to moles of #"Ca"_3("PO"_4)_2#
#"moles of Mg"("NO"_3)_2 stackrelcolor(blue)("molar ratio"color(white)(Xl))( →) "moles of Ca"_3("PO"_4)_2#
(d) Finally, we can use the molar mass to convert the moles of #"Ca"_3("PO"_4)_2# to mass of #"Ca"_3("PO"_4)_2#.
Our complete strategy is:
#"Mass of Mg"("NO"_3)_2stackrelcolor (blue)("molar mass"color(white)(Xl))(→) "moles of Mg"("NO"_3)_2stackrelcolor (blue)("molar ratio"color(white)(Xl))( →) "moles of Ca"_3("PO"_4)_2 stackrelcolor (blue)("molar mass"color(white)(Xl))(→) "mass of Ca"_3("PO"_4)_2#
Solution
(a) The balanced equation is
#"3Mg"("NO"_3)_2 + "Ca"_3("PO"_4)_2 → "Mg"_3("PO"_4)_2 + "3Ca"("NO"_3)_2#
(b) Calculate moles of #"Mg"("NO"_3)_2#
#100 color(red)(cancel(color(black)("g Mg"("NO"_3)_2))) × ("1 mol Mg"("NO"_3)_2)/(148.31 color(red)(cancel(color(black)("g Mg"("NO"_3)_2)))) = "0.6743 mol Mg"("NO"_3)_2#
(c) Calculate moles of #"Ca"_3("PO"_4)_2#
The molar ratio of #"Ca"_3("PO"_4)_2# to #"Mg"("NO"_3)_2# is #("1 mol Ca"_3("PO"_4)_2)/("3 mol Mg"("NO"_3)_2)"#
#"Moles of Ca"_3("PO"_4)_2 = 0.6743 color(red)(cancel(color(black)("mol Mg"("NO"_3)_2))) × ("1 mol Ca"_3("PO"_4)_2)/(3 color(red)(cancel(color(black)("mol Mg"("NO"_3)_2)))) = "0.2248 mol Ca"_3("PO"_4)_2#
(d) Calculate the mass of #"Ca"_3("PO"_4)_2#
#0.2248 color(red)(cancel(color(black)("mol Ca"_3("PO"_4)_2))) × ("310.18 g Ca"_3("PO"_4)_2)/(1 color(red)(cancel(color(black)("mol Ca"_3("PO"_4)_2)))) = "69.7 g Ca"_3("PO"_4)_2 #
Answer: #"69.7 g Ca"_3("PO"_4)_2 # will react.
