# Question #3c9f7

##### 1 Answer

#### Explanation:

You don't even need the balanced chemical equation for this one, all you need to use to solve this problem is the Law of Conservation of Mass.

So, the first sample produced

#"mass MgF"_2 = "1.64 kg" + "2.56 kg" = "4.20 kg"#

Out of this,

#(1.64 cancel("kg"))/(4.20 cancel("kg")) * 100% = 39%# by mass is magnesium

and

#(2.56 cancel("kg"))/(4.20 cancel("kg")) * 100% = 61%# by mass is fluorine.

Since *the same compound undergoes decomposition in the second reaction*, the same proportions by mass will be true for both magnesium and fluorine.

So, the second sample produces

#1.29 cancel("kg Mg") * "100 g MgF"_2/(39cancel("kg Mg")) = "3.31 kg MgF"_2#

is present in the second sample, which means that

#3.31 cancel("kg MgF"_2) * "61 kg F"/(100 cancel("kg MgF"_2)) = "2.02 kg F"#

will be produced this time around. As a side, keep in mind that the reaction will produce fluorine gas,

Verify that the mass is conserved by

#"3.31 kg" = "1.29 kg" + "2.02 kg"#

This confirms the calculations for fluorine and for the magnesium fluoride.

Finally, to convert this to *grams*, use the fact that

#"1 kg = 1000 g"#

You should end up with

#2.02 cancel("kg") * "1000 g"/(1cancel("kg")) = "2020 g" -># rounded to threesig figs