Question

1) What is the mass of silver that can be prepared from 1.40 g of copper metal?
Cu(s)+2AgNO3(aq)→Cu(NO3)2(aq)+2Ag(s)

2a) How many moles of chlorine gas react with 0.160 mol of metallic iron?

2b) How many moles of iron(III) chloride are produced?

Answer 1

Balanced equation:
Cu(s) + 2 AgNO₃(aq) ===> Cu(NO₃)₂(aq) + 2 Ag(s)
Reaction type: single replacement

1.40 g of copper metal = 1.4 / 63.54 = 0.0220 Moles

Moles of SIlver to be prepared =  0.0440 Moles

Mass of silver to be prepared =  0.0440 x 107.86 = 4.752 gm

Quesiton 2a

Balanced equation:
2 Fe + 3 Cl₂ ===> 2 FeCl₃
Reaction type: synthesis

Moles of metallic ion = 0.16 Moles

Moles of Cl2 reacted = 0.24 Moles

because according to the equation 2 moles of Fe needs 3 moles of Cl2 gas.

Moles FeCl3 to be produced = 0.16 Moles

because according to the equation 1 moles of Fe will give 1 mole FeCl3.