Question #713ec
1 Answer
Your reaction will produce 20.7 g of lead metal.
Explanation:
Start with the balanced chemical equation for this redox reaction
#3PbO_((s)) + 2NH_(3(g)) -> N_(2(g)) + 3Pb_((s)) + 3H_2O_((g))#
Notice that you have a
SImply put, for every mole of lead (II) oxide you will need
Use the molar masses of the two compounds 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"#
and
#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 limiting reagent 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
#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")#