Question bd620

Sep 19, 2017

$\text{426 g}$

Explanation:

Notice that the chemical formula of sodium chlorate tells you that $1$ mole of this compound contains

• $1$ mole of sodium atoms, $1 \times \text{Na}$
• $1$ mole of chlorine atoms, $1 \times \text{Cl}$
• $3$ moles of oxygen atoms, $3 \times \text{O}$

This means that if you take a sample of sodium chlorate that contains exactly $1$ mole of sodium chlorate, you can use the molar masses of sodium chlorate and of its constituent elements to say that you have

overbrace("106.4412 g")^(color(blue)("1 mole of NaClO"_3)) = overbrace("22.99 g")^(color(blue)("1 mole of Na")) + overbrace("35.453 g")^(color(blue)("1 mole of Cl")) + overbrace(3 xx "15.9994 g")^(color(blue)("3 moles of O"))#

This tells you that for every $\text{106.4412 g}$ of sodium chlorate, the equivalent of $1$ mole of this compound, you get $\text{35.453 g}$ of chlorine atoms, the equivalent of $1$ mole of chlorine.

Therefore, you can say that your sample will have a mass of

$142 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g Cl"))) * "106.4412 g NaClO"_3/(35.453color(red)(cancel(color(black)("g Cl")))) = color(darkgreen)(ul(color(black)("426 g NaClO}}_{3}}}}$

The answer is rounded to three sig figs.