Calculate the mass of AgCl that can be prepared from 155 g of NiCl2 and sufficient (i.e. excess) AgNO3.?

Question

Calculate the mass of AgCl that can be prepared from 155 g of NiCl2 and sufficient (i.e. excess) AgNO3.

A. 85.5 g
B. 171 g
C. 120 g
D. 239 g
E. 342 g

Answer 1

First, you must write the balanced equation for the reaction, which is:

NiCl2 + 2AgNO3 --> 2AgCl + Ni(NO3)2

Then, you can use the factor label method to determine how many grams of AgCl will be produced if 155 grams of NiCl2 is reacted with excess AgNO3.

The answer will be E, 342 g.

Answer 2

All these problems work the same way. You start with what you are given, and use "conversion factors" to get to what you want. The roadmap always looks a bit like this:

g of known reagent --> moles known reagent --> moles unknown reagent --> g unknown reagent

In the conversion factors, the thing on top has got to be worth exactly the same as the thing on the bottom, and if you cancel out the units as you go that will keep you on the right road. This always works.

You always need a balanced equation to go from moles of one substance to moles of another.

In this case, we have

155 g NiCl2 x (1 mole NiCl2)/(molar mass of NiCl2) [to get to moles NiCl2]

x (2 mol AgCl)/(1 mol NiCl2) [because each NiCl2 gives you two AgCl; you are now in mol AgCl]

x (molar mass AgCl)/(1 mol AgCl) [to give you g AgCl,which is what you want].

Check that the units cancel as they should. Then shove in the numbers and do the arithmetic. In this case, since it is multiple choce you can probably pick out the right answer in your head.

That's all!

Answer 3

Molar Masses: Ag = 107.9g, Ni= 58.7g, Cl= 35.5g
So 1mol of NiCl2 = 58.7 + ( 35.5 x 2) = 129.7g
Thus 155g of NiCl2 = 155/129.7 = 1.195Mol
So 1.195 x 2 Mol of AgCl can be prepared = 2.39 Mole
1Mol of AgCl = 107.9 + 35.5 = 143.4g
Thus 2.39 x 143.4 = 342.7g of AgCl can be prepared.
So E is Correct 342 g

Answer 4

the answer is E...if you want a detailed solution just let me know :)