How would you solve this Molarity Problem?

Question

I do not understand how to do this question.

How would you go about doing it?

A.) How many milliliters of 0.248 M HCl are needed to react with 1.36 g of Zinc to produce hydrogen?

B.) What mass of precipitate will be isolated from the reaction of 200mL of 0.50 M potassium iodide and 100mL of 1.0 M lead (II) nitrate?

Thank you so much!

Answer 1

Zn + 2HCl -----> ZnCl2     +   H2

atomic mass (Zn)=65.37g/mol

molecular mass(HCl)=1+35.5

=36.5

65.37 g of zinc is reacting with 73g of HCl

1.36 g of zinc will react with =73*1.36/65.37

=1.51 g of HCl

36.5 g of HCl= 1 mol

1.51 g of HCl=0.04mol

Molarity of HCl=0.248M

let volume be V

molarity = number of moles of solute / volume of solution in litres

0.248=0.04/V

V=0.04/0.248

=0.16 litres

1 litres = 1000 millilitres

0.16 litres=1000*0.16

=160millilitres

b)

for potassium iodide

Volume=200 ml =200*10^ -3 litres

molarity(M)=0.50

molecular mass of KI=39+127=166 g/mol

number of moles of solute=amount in grams/molecular mass

molarity=number of moles of solute / volume of solution in litres

amount in grams=0.50*166*200*10^ -3

=16.6 g

for lead nitrate

Pb(NO3)2

molecular mass of lead nitrate=207+2*(14+3*16)

=331 g/mol

using same formula we can calculate amount in grams

i.e. =33.1 g

2KI(aq) + Pb(NO3)2(aq) -> 2KNO3(aq) + PbI(s)

332 grams of KI reacting with 331 g of Pb(NO3)2

16.6 grams of KI will react with =331*16.6/332

=16.55 grams of Pb(NO3)2

but amount of Pb(NO3)2 present is 33.1 g

so KI is limiting reagent which limits the formation of products

2KI(aq) + Pb(NO3)2(aq) -> 2KNO3(aq) + PbI(s)

molecular mass of PbI=207+127=334 g/mol

332 grams of KI is forming  334 g of PbI

16.6 grams of KI will form = (334*16.6)/332

=16.7 g of PbI

16.7 g of precipitate will be isolated from reaction

Answer 2

a) Zn + 2H+ -> Zn++ + H2

moles Zn = 1.36/ at wt Zn

moles H+ needed = 2*moles Zn

ml of HCl = 1000 ml/liter * moles Zn/0.248 moles H+/liter

b) 2KI + Pb(NO3)2 -> PbI2

grams PbI2 = [lesser of (1/2 * moles of KI) or (moles of Pb)] * MW PbI2

moles KI = 0.200 * 0.50 = 0.1 mole I- available; so can generate only 0.05 moles PbI2

moles Pb = 0.100 * 1 = 0.1

Thus KI is the limiting factor.

I leave the other math to you

Answer 3

A) Zn + 2 HCl >> ZnCl2 + H2

Moles Zn = 1.36 g /65.409 g/mol = 0.0208

Moles HCl needed = 2 x 0.0208 =0.0416

V = moles / M = 0.0416 / 0.248 =0.168 L => 168 mL

2KI + Pb(NO3)2 >> PbI2 + 2 KNO3

Moles KI = 0.200 L x 0.50 M =0.10

Moles Pb(NO3)2 = 0.100 L x 1.0 M = 0.100

the ratio is 2 : 1 so KI is the limiting reactant

we will get 0.100/2 =0.050 moles of PbI2

Mass PbI2 = 0.050 mol x 461.05 g/mol =23.1 g

Answer 4

i) V*.248=2*1.36/65

V=2*1.36/65*.248=0.1687L =168.7 mL