Answer:
(a). mass fraction = 32.32%.
(b). the feed rate = 257 L/h
Explanation:
from the question it was given that the First aqueous soln contained 20.0 wt% H₂SO₄ with a specific gravity (s₁) of 1.139.
Also, the second aqueous soln contained 60.0 wt % H₂SO₄ with a specific gravity (s₂) of 1.498.
The combined solution when mixed produces a 4M soln with specific gravity (sm) = 1.213
let us profer solution for the question stated.
(a). we are asked to calculate the mass fraction of sulphuric acid in the product solution, but first we must show the relationship connecting it to the molarity.
Molarity = moles of acid/volume of solution (L) = weight of acid/molecular weight × volume of solution (L) ....................(1)
given our specific gravity of mixed soln (sm) = 1.1213, we would calculate the specific gravity
specific gravity = density of solution (ρsol) / density of water (ρwater)
this gives ρsoln = 1.1213 × ρwater...............(2)
ρsoln = 1213 × 1000 (kg/m³) where ρwater = 1000(kg/m³)
we know that ρsol = weight of soln/voulme of soln
therefore volume of soln = weight of soln / ρsol ................(3)
Recall from equation (1), substituting equation (3) we have
Molarity = weight of acid × ρsol / molecular weight × weight of sol
4 = weight of acid × 1213 / 98 × weight of sol
taking terms to the other side,
4 × 98 / 1213 = weight of acid / weight of soln
0.3232 = weight of acid / weight of soln
∴ weight of acid / weight of soln (in percent) = 32.32 %
so the mass fraction of the sulphuric acid in product solution becomes 32.32%.
(b). Given that we are asked to find the rate of feed of 60% solution to produce 1250 Kg/h of the product,
we have that wt of acid = 0.3232 × wt of sol
where the weight of sol = 1250 Kg/h
∴ we have that wt of acid = 0.3232 × 1250 (Kg/h) = 404 Kg/h
taken overall balance of solution as
wt₁ + wt₂ = 1250 Kg/h
where wt₁ = wt of the first solution
and wt₂ = wt of the second solution.
Taken mass balance of acid
wt₁(0.2) + wt₂(0.6) = 404 Kg/h..........(1)
solving simultaneously we have
wt₁ = 865 Kg/h and wt₂ = 385 Kg/h
calculating the feed rate at 60% i.e at second solution, we have
volume of sol = wt of second sol / ρsoln
volume of sol = 385 / 1498 = 0.257 m₃/h × 1000 L/m³
where ρ = 1.498 × 1000 Kg/m³ = 1498 Kg/m³
∴ the feed rate of the solution = 385 / 1498 = 0.257 m₃/h × 1000 L/m³
the feed rate of the solution = 257 L/h
cheers i hope this helps
