Question

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Fall 2019 NAME: Test No. 8. (8) What volume of 0.562 M HCI is required to titrate 36.8 ml. of 0.386 M Ba(OH)2? 9. (12') When you mix 25.0 mL of 0.225 M Pb(NO3)2 with 35.0 mL of 0.148 M KaS, a precipitate will form. (aq)+ K2S (aq) - PbS (s)+2 KNO (aq) Pb(NOs)2 How many moles of precipitate will form? b. What is the theoretical yield of the reaction? How many grams of precipitate can be collected after the experiment if the percent yield is 75.0% ? а. c.

Answer 1

8) First see balance reaction between Ba(OH)₂ and HCl

Ba(OH)₂ + 2HCl  $\LARGE \rightarrow$ BaCl₂ + 2H₂O

Here, one mole of barium hydroxide react with two moles of HCl gives barium chloride and water as product.

We have to calculate moles of Ba(OH₂) ,because we know amount and strength of barium hydroxide.

# Amount of barium hydroxide(V₁) = 36.8mL

#strength of Barium hydroxide (M₁)= 0.386M

# strength of HCl( M₂) = 0.562 M

#Volume of HCl required = ?

We can use,

M₁ V₁  = M₂V₂ Formula

By putting values into equation -

0.386 M x 36.8 mL = 0.562 M x V₂

So ,

V₂= (0.386 x 36.8 )/(0.526)

= 25.27mL

But 2.0 mole of HCl solution require for one mole of barium hydroxide, so total HCl required for complete reaction ,

= 25.27 mL x 2

= 50 .55 mL of 0.562M HCl solution required .

9)solution-

We know definition of Molarity -

Number of moles per liter of solution is called as Molarity .

From balance reaction it is clear that one mole of lead nitrate react with potassium sulfide to give one mole of lead sulfide , so we have to calculate number of moles from strength and Molarity .

#Molarity (M) = Number of moles (n) / per liter of solution (L)

By rearranging above formula ,

#Number of moles (n) = Molarity (M) x volume of solution in liter

Let's given strength in terms of Molarity and quantity in liter

# lead nitrate =25 mL =0.025 Liter

# strength of lead nitrate =0.225M

# Molar mass of lead nitrate =331.2 g/mol

So, number of moles (n)= 0.225 M x 0.025 Lit

= 0.005625 moles

We have to know about limiting reagent by isolation of limiting reactant.

# Molarity of K₂ S =  0.148 M

# amount of K₂S = 35 mL = 0.035 mL

Moles of K₂S = 0.148 M x 0.035 ML

= 0.00518 moles of K₂S

So moles of K₂S less than moles of lead nitrate, so moles of K₂ S is limiting reagent .(0.00518 mol $\LARGE >$0.005625 mol) ,so moles of K₂S decided the amount of product .

a) number of moles of product fotmed= 0.00518 mol(Limiting reagent )

b) molar mass of lead  sulfide =239.3 g/mol

100% theoretical yield of lead sulfide = moles of limitating reagents x molar mass of lead sulfide

= 0.00518 mol x 239.3 g/mol

= 1.24 g of PbS will be formed .

c) how many grams of lead sulfide formed reaction has 75 %yiled

75 % yield =( 1.24 x75 )/100

=0.930 g PbS will be formed .