Question

1. A student reacts 1.60 ml of 15.7M HNO3 with 0.5000g of copper metal.  Determine the limiting reactant and the amount of excess reactant (in moles) left over, assuming the reaction goes to completion.

2. A student dissolves 5.0000g CoCl2*6H2O(s) in water, reactsit with excess C2H8N2 and obtains 4.0000g of solid [Co(C2H8N2)3]Cl2 after filtering and drying the product. What is his percent yield, assuming there are no other cobalt‐containing reactants or products?

3. A student was given 15.0 ml of a 22% by volume solution of chloroform in water. The density of chloroform, CHCl3, is 1.48 g/ml.  How many moles of chloroform did he have?

Answer 1

moles of NH3 in 1.6v ml of 15.7 M =15.7*1.6/1000 =0.02512 moles

Mass of Coppier =0.5 gms ,moles of copper = 0.5/63.5 =0.007874

The reaction can be represented as

Cu(s) + 4HNO3(aq) ——> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

Molar ratio of reactants as per the reaction = 1:4

Molar ratios considered =0.007874 :0.02512 = 1:3.2

limiting reactant is HNO3

Moles of Cu that can react with 0.02512 moles of HNO3 is 0.02512/4= 0.00628 moles

Moles actually taken =0.007874

moles excess =0.007874-0.00628=0.001594

% excess =(0.01594/0.00628)*100=25.38%

2.Molecular weight of CoCl2.6H2O= 59+ 71+108= 238

Mole of COCl2.6H2O= 5/238 =0.021 moles

Moles of CO =0.021 moles

Moles of CO in 4gns of Co(C2H8N2)3Cl2 ( molecular wieght = 59+ 3*(24+8+28)+ 71 =59+ 180+71=320

=4/320 =0.0125 moles

moles converted = 0.021-0.0125= 0.0085

Yiled of Co = (0.0085/ 0.021)*100= 68%

3. Volume of solution =100 ml

Chloroform =22 ml   volume of water =88 ml

15 ml of this solution contains 22*15/100 =3.3 ml of CHCl3

Density of CHCl3= 1.48 g/ml

mass of CHCl3 =3.3*1.48 = 4.9 gms , moles of CHCl3 =4.9/ 119 =0.041176