Question 8aae5

Sep 23, 2015

$\text{21 g}$

Explanation:

The quckest way of finding the mass of oxygen found in that many moles of sodium carbonate decahydrate, $\text{Na"_2"CO"_3 * 10"H"_2"O}$, is to find the number of moles of oxygen you get in that many moles of decahydrate.

Once you know how many moles of oxygen you have, you can use oxygen's molar mass to find how many grams you have.

You can find the number of moles of oxygen by inspecting the chemical formula of the decahydrate.

Every mole of the anhydrous salt, ${\text{Na"_2"CO}}_{3}$, contains

• two moles of sodium
• one mole of carbon
• three moles of oxygen

Every mole of water, $\text{H"_2"O}$, contains

• two moles of hydrogen
• one mole of oxygen

Since every mole of decahydrate contains 10 moles of water, you know that one mole of decahydrate will contain

underbrace("3 moles O")_(color(blue)("from the anhydrous salt")) + 10 xx overbrace("1 mole O")^(color(orange)("from the water")) = "13 moles O"

This means that 1 mole of $\text{Na"_2"CO"_3 * 10"H"_2"O}$ contains 13 moles of oxygen.

Since you have 0.10 moles of $\text{Na"_2"CO"_3 * 10"H"_2"O}$, you have

0.10color(red)(cancel(color(black)("moles decahydrate"))) * "13 moles O"/(1color(red)(cancel(color(black)("mole decahydrate")))) = "1.3 moles O"

Oxygen's molar mass is euqal to $\text{15.9994 g/mol}$, which means that 0.10 moles of sodium carbonate decahydrate will contain

1.3color(red)(cancel(color(black)("moles O"))) * "15.9994 g"/(1color(red)(cancel(color(black)("mole oxygen")))) = "20.799 g O"#

Rounded to two sig figs, the number of sig figs you gave for the moles of sodium carbonate decahydrate, the answer will be

$\textcolor{g r e e n}{\text{21 g oxygen}}$