# Help with these three questions about mass and concentrations?

## 1) If $\text{450 kg}$ of ${\text{SO}}_{2}$ reacts with coke (${\text{CS}}_{2}$) with 82% consumption, what mass of coke is formed? 2) In a $\text{29 g}$ mixture of $\text{BaO}$ and $\text{CaO}$, $\text{100.8 mL}$ of $\text{6 M HCl}$ neutralizes the mixture. What is the $\text{%w/w BaO}$? 3) If the true specific gravity of ${\text{H"_2"SO}}_{4} \left(l\right)$ is $1.2$, then if we have a 27%"w/w" solution of ${\text{H"_2"SO}}_{4}$, what is its molarity?

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 $S {O}_{2}$ produces $1 m o l \text{ } C {S}_{2}$

or $2 \times 64 g$ $S {O}_{2}$ produces $76 g \text{ } C {S}_{2}$

or $2 \times 64 k g$ $S {O}_{2}$ produces $76 k g \text{ } C {S}_{2}$

$369 k g S {O}_{2}$ will produce $\frac{76}{128} \times 369 \approx 219.09 k g \text{ } C {S}_{2}$

Q-2

Let there be $x$ mol $B a O$ and $y$ mol $C a O$ in 29 g mixture.

The molar masses of

$B a O \to 153 g \text{/} m o l$

$C a O \to 56 g \text{/} m o l$

So by the condition

$153 x + 56 y = 29. \ldots . \left[1\right]$

Again $100.8 m l \mathmr{and} 0.1008 L$ $6 M$ HCl solution is required to neutralize the mixture.
So amount of HCl required is $0.1008 \times 6$ mol

Now by stoichiometry of the given reaction we see that number of moles of both $B a O \mathmr{and} C a O$ are $\frac{1}{2}$ of the number of moles of $H C l$ So total number of moles of $B a O \mathmr{and} C a O$ will be $\frac{1}{2} \times 0.1008 \times 6 = 0.3024$ mol

Hence $x + y = 0.3024 \ldots \ldots . \left[2\right]$

Multiplying  by 56 and subtracting the resulting equation from  we get

$\left(153 - 56\right) x = \left(29 - 56 \times 0.3024\right)$

$\implies x = \frac{29 - 56 \times 0.3024}{97} \approx 0.124 m o l$

So mass of $B a O$ in 29g mixture $153 \times 0.124 g$

So percentage of $B a O$ in the mixture

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

Q-3

Given w/w% of H_2SO_4=27

So 100g solution contains 27g or $\frac{27}{98}$ mol ${H}_{2} S {O}_{4}$

Or $\frac{100}{1.2}$ mL solution contains $\frac{27}{98}$ mol ${H}_{2} S {O}_{4}$

Hence 1000mL solution contains $\frac{\frac{27}{98}}{\frac{100}{1.2}} \times 1000$ mol ${H}_{2} S {O}_{4}$

$= \frac{270}{98} \cdot 1.2 m o l$ ${H}_{2} S {O}_{4} = 3.30 m o l \text{/} L$

Hence molarity of the solution is $3.30 M m$

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 $\text{1 g/mL}$, then the specific gravity is numerically the density of the solution, i.e.

$\text{SG} = \frac{{\rho}_{{H}_{2} S {O}_{4} \left(a q\right)}}{{\rho}_{{H}_{2} O \left(l\right)}} = 1.2$

$\implies {\rho}_{{H}_{2} S {O}_{4} \left(a q\right)} = 1.2 \cdot \text{1 g/mL" = "1.2 g/mL}$

Then, we are given:

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

Since molarity is defined as

$\text{mols solute"/"L soln}$,

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")

$=$ $\text{0.833 L soln}$

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))

$= {\text{2.753 mols H"_2"SO}}_{4}$

This means we have a molarity of:

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

$=$ $\textcolor{b l u e}{\text{3.30 M}}$

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