Help with these three questions about mass and concentrations?

Question

#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?

Ellie Andrews · Accepted Answer

#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"#

Next, we receive:

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

As the definition of molarity is

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

we require the solution's volume. With the solution's specific gravity, we get:

#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"#

Next, we require the solute's mols:

#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 indicates that our molarity is:

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

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

The rest is correct; either the answer key is wrong or there's a problem with my density.

Peyton Adams · Answer

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 the #BaOto153g"/"mol#  #CaOto56g"/"mol#  Thus, based on the circumstance #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]# After subtracting the resultant equation from [1] and multiplying [2] by 56, we obtain #(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#

Olivia Anton · Answer

undefined Certainly, please provide the three questions about mass and concentrations that you need help with.