Help with these three questions about mass and concentrations?

#1)# If #"450 kg"# of #"SO"_2# reacts with coke (#"CS"_2#) with #82%# consumption, what mass of coke is formed?
#2)# In a #"29 g"# mixture of #"BaO"# and #"CaO"#, #"100.8 mL"# of #"6 M HCl"# neutralizes the mixture. What is the #"%w/w BaO"#?
#3)# If the true specific gravity of #"H"_2"SO"_4(l)# is #1.2#, then if we have a #27%"w/w"# solution of #"H"_2"SO"_4#, what is its molarity?

2 Answers
Jun 9, 2017

Q-1

As per given condition #82% of 450kg SO_2=369kg SO_2# has taken part in the reaction with coke . By the given equation we see

2 moles #SO_2# produces #1mol" "CS_2#

or #2xx64g# #SO_2# produces #76g" "CS_2#

or #2xx64kg# #SO_2# produces #76kg" "CS_2#

#369kg SO_2# will produce #76/128xx369~~219.09kg" "CS_2#

Q-2

Let there be #x# mol #BaO# and #y# mol #CaO # in 29 g mixture.

The molar masses of

#BaOto153g"/"mol#

#CaOto56g"/"mol#

So by the condition

#153x+56y=29.....[1]#

Again #100.8mlor0.1008L # #6M# HCl solution is required to neutralize the mixture.
So amount of HCl required is #0.1008xx6# mol

Now by stoichiometry of the given reaction we see that number of moles of both #BaO and CaO# are #1/2# of the number of moles of #HCl# So total number of moles of #BaO and CaO# will be #1/2xx0.1008xx6=0.3024# mol

Hence #x+y=0.3024.......[2]#

Multiplying [2] by 56 and subtracting the resulting equation from [1] we get

#(153-56)x=(29-56xx0.3024)#

#=>x=(29-56xx0.3024)/97~~0.124mol#

So mass of #BaO# in 29g mixture #153xx0.124g#

So percentage of #BaO# in the mixture

#=(153xx0.124g)/29xx100%~~65.65%#

Q-3

Given #w/w% of H_2SO_4=27#

So 100g solution contains 27g or #27/98# mol #H_2SO_4#

Or #100/1.2# mL solution contains #27/98# mol #H_2SO_4#

Hence 1000mL solution contains #(27/98)/(100/1.2)xx1000# mol #H_2SO_4#

#=270/98*1.2mol# #H_2SO_4=3.30mol"/"L#

Hence molarity of the solution is #3.30Mm#

Jun 9, 2017

#3)#

If the true specific gravity of the solution is #1.2#, we expect that it is with respect to the density of water because the solution is a liquid. If we assume the density of water is #"1 g/mL"#, then the specific gravity is numerically the density of the solution, i.e.

#"SG" = (rho_(H_2SO_4(aq)))/(rho_(H_2O(l))) = 1.2#

#=> rho_(H_2SO_4(aq)) = 1.2 cdot "1 g/mL" = "1.2 g/mL"#

Then, we are given:

#%"w/w" = ("270 g H"_2"SO"_4)/("1000 g soln")#

Since molarity is defined as

#"mols solute"/"L soln"#,

we need the volume of the solution. Using the specific gravity of the solution, we have:

#V_(sol n) = 1000 cancel"g soln" xx cancel"1 mL"/(1.2 cancel"g soln") xx "1 L"/(1000 cancel"mL")#

#=# #"0.833 L soln"#

We next need the mols of solute:

#270 cancel("g H"_2"SO"_4) xx "1 mol"/(98.079 cancel("g H"_2"SO"_4))#

#= "2.753 mols H"_2"SO"_4#

This means we have a molarity of:

#color(blue)(["H"_2"SO"_4]) = ("2.753 mols H"_2"SO"_4)/("0.833 L soln")#

#=# #color(blue)("3.30 M")#

Either the answer key is incorrect or there's something off with my density. The rest is correct.