The chemical equation and the molar masses are
M_text(r): color(white)(mmml)286.14color(white)(mmmmmmmmmmmmm)18.02Mr:mmml286.14mmmmmmmmmmmmm18.02
color(white)(mmm)"Na"_2"CO"_3·"10H"_2"O" → "Na"_2"CO"_3·"H"_2"O" + "9H"_2"O"mmmNa2CO3⋅10H2O→Na2CO3⋅H2O+9H2O
Step 1. Calculate the moles of "Na"_2"CO"_3·"10H"_2"O"Na2CO3⋅10H2O.
"Moles of Na"_2"CO"_3·"10H"_2"O"Moles of Na2CO3⋅10H2O
= 100.0 color(red)(cancel(color(black)("g Na"_2"CO"_3·"10H"_2"O"))) × ("1 mol Na"_2"CO"_3·"10H"_2"O")/(286.14 color(red)(cancel(color(black)("g Na"_2"CO"_3·"10H"_2"O")))) = "0.3495 mol Na"_2"CO"_3·"10H"_2"O"
Step 2. Calculate the moles of water released.
"Moles of H"_2"O" = 0.3495 color(red)(cancel(color(black)("mol Na"_2"CO"_3·"10H"_2"O"))) × ("9 mol H"_2"O")/(1 color(red)(cancel(color(black)("mol Na"_2"CO"_3·"10H"_2"O")))) = "3.145 mol H"_2"O"
Step 3. Calculate the mass of water released.
"Mass of water" = 3.145 color(red)(cancel(color(black)("mol H"_2"O"))) × ("18.02 g H"_2"O")/(1 color(red)(cancel(color(black)("mol H"_2"O")))) = "56.68 g H"_2"O"
