Question #42189

Start by taking a look at the balanced chemical equation for this single replacement reaction

#"Zn"_text((s]) + color(red)(2)"Au"("NO"_3)_text(3(aq]) -> 2"Au"_text((s]) + 3"Zn"("NO"_3)_text(2(aq])#

The key to this problem is the #1:color(red)(2)# mole ratio that exists between zinc metal and gold(III) nitrate. This mole ratio tells you that the reaction will consume #color(red)(2)# moles of gold(III) nitrate for every #1# mole of zinc metal that takes part in the reaction.

Your goal now is to figure out exactly how many moles of each reactant you're mixing. To do that, use zinc's molar mass and the molarity and volume of the gold(III) nitrate solution.

For zinc metal, you'll have

#1.00color(red)(cancel(color(black)("g"))) * "1 mole Zn"/(65.38color(red)(cancel(color(black)("g")))) = "0.01530 moles Zn"#

For gold(III) nitrate, you'll have

#color(blue)(c = n/V implies n = c * V)#

#n = "0.0500 M" * 50.0 * 10^(-3)"L" = "0.00250 moles Au"("NO"_3)_3#

It looks like you don't have enough moles of gold(III) nitrate to allow for all the moles of zinc to react. More specifically, the #0.00250# moles of gold(III) nitrate will only allow

#0.00250color(red)(cancel(color(black)("moles Au"("NO"_3)_3))) * "1 mole Zn"/(color(red)(2)color(red)(cancel(color(black)("moles Au"("NO"_3)_3)))) = "0.00125 moles Zn"#

out of the total #0.01530# moles of zinc to take part in the reaction. Therefore, you can say that gold(III) nitrate will be the limiting reagent in this reaction.

Now, the #color(red)(2):2# mole ratio that exists between gold(III) nitrate and gold tells you that the reaction will produce the same number of moles of the latter as you have moles of the former.

Since #0.00250# moles of gold(III) nitrate take part in the reaction, it follows that #0.00250# moles of gold will be produced.

To determine how many grams of gold would contain this many moles, use gold's molar mass

#0.00250color(red)(cancel(color(black)("moles Au"))) * "196.97 g"/(1color(red)(cancel(color(black)("mole Au")))) = color(green)("0.492 g Au")#