Question

Question #bd620

Answer 1

`"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 xx "Na"`
• `1` mole of chlorine atoms, `1 xx "Cl"`
• `3` moles of oxygen atoms, `3 xx "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 `"106.4412 g"` of sodium chlorate, the equivalent of `1` mole of this compound, you get `"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 color(red)(cancel(color(black)("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.