# Question 88b4c

Mar 13, 2017

The decrease in mass will be 56.68 g.

#### Explanation:

The chemical equation and the molar masses are

${M}_{\textrm{r}} : \textcolor{w h i t e}{m m m l} 286.14 \textcolor{w h i t e}{m m m m m m m m m m m m m} 18.02$
$\textcolor{w h i t e}{m m m} \text{Na"_2"CO"_3·"10H"_2"O" → "Na"_2"CO"_3·"H"_2"O" + "9H"_2"O}$

Step 1. Calculate the moles of $\text{Na"_2"CO"_3·"10H"_2"O}$.

$\text{Moles of Na"_2"CO"_3·"10H"_2"O}$

= 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.

$\text{Moles of H"_2"O" = 0.3495 color(red)(cancel(color(black)("mol Na"_2"CO"_3·"10H"_2"O"))) × ("9 mol H"_2"O")/(1 color(red)(cancel(color(black)("mol Na"_2"CO"_3·"10H"_2"O")))) = "3.145 mol H"_2"O}$

Step 3. Calculate the mass of water released.

$\text{Mass of water" = 3.145 color(red)(cancel(color(black)("mol H"_2"O"))) × ("18.02 g H"_2"O")/(1 color(red)(cancel(color(black)("mol H"_2"O")))) = "56.68 g H"_2"O}$