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
Q-1
As per given condition
2 moles
or
or
Q-2
Let there be
The molar masses of
So by the condition
Again
So amount of HCl required is
Now by stoichiometry of the given reaction we see that number of moles of both
Hence
Multiplying [2] by 56 and subtracting the resulting equation from [1] we get
So mass of
So percentage of
Q-3
Given
So 100g solution contains 27g or
Or
Hence 1000mL solution contains
Hence molarity of the solution is
If the true specific gravity of the solution is
#"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.