#"ZnSO"_4 + "Li"_2"CO"_3"##rarr##"ZnCO"_3 + "Li"_2"SO"_4#
The process used to answer this question will be:
#color(red)("given mass Li"_2"CO"_3"##rarr##color(blue)("mol Li"_2"CO"_3"##rarr##color(green)("mol ZnCO"_3"##rarr##color(purple)("mass ZnCO"_3"#
The molar masses of lithium carbonate and zinc sulfate is needed.
#"Li"_2"CO"_3:##"73.888 g/mol"#
https://pubchem.ncbi.nlm.nih.gov/compound/11125
#"ZnCO"_3:##"125.388 g/mol"#
https://www.ncbi.nlm.nih.gov/pccompound?term=%22ZINC+CARBONATE%22
#color(red)("Given Mass Li"_2"CO"_3"##rarr##color(blue)("Mol Li"_2"CO"_3"#
Divide the given mass by the molar mass by multiplying the given mass by the inverse of the molar mass.
#38color(red)cancel(color(black)("g Li"_2"CO"_3))xx(1"mol Li"_2"CO"_3)/(73.888color(red)cancel(color(black)("g Li"_2"CO"_3)))="0.51429 mol Li"_2"CO"_3"#
#color(blue)("Mol Li"_2"CO"_3"##rarr####color(green)("Mol ZnCO"_3"#
Multiply the mol lithium carbonate by the mole ratio from the balanced equation so that mol lithium carbonate cancels.
#0.51429color(red)cancel(color(black)("mol Li"_2"CO"_3))xx(1"mol ZnCO"_3)/(1color(red)cancel(color(black)("mol Li"_2"CO"_3)))="0.51429 mol ZnCO"_3"#
#color(green)("Mol ZnCO"_3"##rarr##color(purple)("Mass ZnCO"_3"#
Multiply mol zinc carbonate by its molar mass.
#0.51429color(red)cancel(color(black)("mol ZnCO"_3))xx(125.388"g ZnCO"_3)/(1color(red)cancel(color(black)("mol ZnCO"_3)))="64 g ZnCO"_3"# rounded to two sig figs due to #"38 g"#.
You can write the steps as one equation, like this:
#(color(red)stackrel"given mass Li2CO3"(38color(black)cancel(color(red)("g Li"_2"CO"_2))))xx(color(blue)stackrel"mole Li2CO3"
(1"mol Li"_2"CO"_3)/(color(blue)(73.888color(black)cancel(color(blue)("g Li"_2"CO"_3)))))xx(color(green)stackrel"mole ZnCO3"(1color(green)("mol ZnCO"_3))/(color(green)(1color(black)cancel(color(green)("mol Li"_2"CO"_3)))))xx(color(purple)stackrel"mass ZnCO3"(125.388"g ZnCO"_3)/color(purple)(0.51429color(black)cancel(color(purple)("mol ZnCO"_3))))=color(purple)("64 g ZnCO"_3#
