1.50 g of aluminium reacted with 35.0 cm3 of 1.00 mol dm–3 solution of copper(II) chloride, and 1.98 g of solid product was collected by filtration. Calculate the percentage yield for this reaction.

89.2%

Explanation:

The reaction of Al with CuCl₂ is:

2Al(s) + 3CuCl₂(aq) → 2AlCl₃(aq) + 3Cu(aq)

Assuming the reaction occurs until limiting reactant is over, we must find limiting reactant:

Moles Al -Molar mass: 26.98g/mol-

1.50g Al * (1mol / 26.98g) = 0.0556 moles

Moles CuCl₂:

35cm³ = 0.035L * (1.00mol / L) = 0.035 moles

For a complete reaction of Al there are necessaries:

0.0556 moles Al * (3 moles CuCl₂ / 2 moles Al) = 0.0834 moles of CuCl₂

As there are just 0.035 moles of CuCl₂, CuCl₂ is limiting reactant.

The solid produced is Cu, the theoretical mass of Cu is:

0.035 moles CuCl₂ * (3 moles Cu / 3 moles CuCl₂) = 0.035 moles Cu

0.035 moles Cu * (63.546g / mol) = 2.22g Cu

As actual yield was 1.98g, percentage yield is:

1.98g Cu / 2.22g Cu * 100 =

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

by cartels in other countries, shipped by air, mail, or land. i cant really see your answer choices : /