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Your reaction will produce 20.7 g of lead metal.
Start with the balanced chemical equation for this
##3PbO_((s)) + 2NH_(3(g)) -> N_(2(g)) + 3Pb_((s)) + 3H_2O_((g))##
Notice that you have a ##3:2## between lead (II) oxide and ammonia, which tells you that your reaction will always consume these two reactants in this ratio.
SImply put, for every mole of lead (II) oxide you will need ##2/3## moles of ammonia, which is equivalent to saying that for every mole of ammonia you need ##3/2## moles of lead (II) oxide.
Use the molar masses of the two to determine how many moles you have of each
##22.3color(red)(cancel(color(black)(“g”))) * “1 mole PbO”/(223.2color(red)(cancel(color(black)(“g”)))) ~= “0.100 moles PbO”##
##34.06color(red)(cancel(color(black)(“g”))) * (“1 mole NH”””_3)/(17.03color(red)(cancel(color(black)(“g”)))) = “2.00 moles NH”””_3##
Notice that you have much more moles of ammonia than of lead (II) oxide, which means that not all the ammonia will react. ANother way of saying this is that you have insufficient lead (II) oxide in order to get all the moles of ammonia to react.
Lead (II) oxide will thus act as a and determine how many moles of ammonia actually take part in the reaction.
##0.100color(red)(cancel(color(black)(“moles of PbO”))) * (“3 moles NH”””_3)/(2color(red)(cancel(color(black)(“moles PbO”)))) = “0.150 moles NH”””_3##
This is how much ammonia will react, the rest will be in excess.
Now notice that you have a ##1:1## (##3:3##) mole ratio between lead (II) oxide and lead metal, which means that the reaction will produce the same number of moles of lead metal as you have moles of lead (II) oxide that react.
##0.100color(red)(cancel(color(black)(“moles PbO”))) * “1 mole Pb”/(1color(red)(cancel(color(black)(“mole PbO”)))) = “0.100 moles Pb”##
Now use lead’s molar mass to see how many grams would contain this many moles
##0.100color(red)(cancel(color(black)(“moles”))) * “207.2 g”/(1color(red)(cancel(color(black)(“mole”)))) = color(green)(“20.7 g Pb”)##
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