Question #8e899

1 Answer
Jul 20, 2017

#"3074 g Mg"("NO"_3)_2#


The first thing to do here is to figure out how many moles of magnesium nitrate are produced by the reaction,

You know by looking at the balanced chemical equation

#2"HNO"_ (3(aq)) + "Mg"("OH")_ (2(s)) -> "Mg"("NO"_ 3)_ (2(aq)) + 2"H"_ 2"O"_ ((l))#

that in order to produce #1# mole of magnesium nitrate, the reaction must consume #2# moles of nitric acid and #1# moles of magnesium hydroxide.

Assuming that you have enough magnesium hydroxide to go around, you can say that the reaction will produce

#41.45 color(red)(cancel(color(black)("moles HNO"_3))) * ("1 mole Mg"("NO"_ 3)_ 2)/(2color(red)(cancel(color(black)("moles HNO"_3)))) = "20.725 moles Mg"("NO"_3)_2#

In order to convert this to grams, use the molar mass of magnesium nitrate

#20.725 color(red)(cancel(color(black)("moles Mg"("NO"_3)_2))) * "148.3 g"/(1color(red)(cancel(color(black)("mole Mg"("NO"_3)_2)))) = color(darkgreen)(ul(color(black)("3074 g")))#

The answer is rounded to four sig figs, the number of sig figs you have for the number of moles of nitric acid that take part in the reaction.