Question

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?

Answer 1

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`

Answer 2

`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.