Question #8e899

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.