Question

Using A CLEAR ICE TABLE and 4SIG FIGS
If 1.200 g of copper(II) chloride is dissolved in water with 1.300 g of silver nitrate, determine what is the limiting reagent, how many grams of precipitate should be formed, and how many moles of excess reagent will remain.should be formed, and how many moles of excess reagent will remain.

Answer 1

Start with the balanced equation.
2AgNO3(aq) + CuCl2(aq) → 2AgCl(s) + Cu(NO3)2(aq).

Determine the number of moles of silver nitrate by dividing the given mass = 1.3 / 169.87 = 0.007652 Mole

Determine the number of moles of copper(II) chloride = 1.2 / 134.45 = 0.00892 Moles

Limiting reagent is silver nitrate wjhich will produce 0.007652 Moles silver chloride precipitate

0.007652 Moles silver chloride precipitate = 0.007652 x 143.32 = 1.096 gm

1.096 gm precipitate will be formed

Used amount of CuCl2 = 0.00382 x 134.52 = 0.514 gm

Excess reagent = 1.2 gm - 0.514 gm = 0.686 gm of CuCl2

Moles of excess reagent = 0.686 / 134.45 = 0.00510 Mole CuCl2