Two aqueous sulfuric acid solutions containing 20.0 wt% h2so4

(sg = 1.139) and 60.0 wt% h2so4 (sg = 1.498) are mixed to form a

4.00 molar solution (sg = 1.213). taking 100kg of the 20% feed

solution as a basis, calculate the feed ratio (liters 20%

solution/liter 60% solution).

The feed ratio (liters 20%  solution/liter 60% solution) is 3,08

Explanation:

In this problem you have a 20,0 wt% H₂SO₄ and a 60,0 wt% H₂SO₄ solutions.

100 kg of 20% solution are 100kg/1,139 kg/L = 87,8 L

100kg×20wt% = 20 kg H₂SO₄. In moles:

20 kg H₂SO₄ × (1 kmol/98,08 kg) = 0,2039 kmol H₂SO₄≡ 203,9 mol

The final molarity 4,00M comes from:

(1)

Where X moles and Y liters comes from 60,0 wt% H₂SO₄

100 L of 60,0 wt% H₂SO₄ are:

100L××× = 0,9164 kmolH₂SO₄ ≡ 916,4 moles

That means:

X/Y = 916,4/100 = 9,164 (2)

Replacing (2) in (1):

Y(liters of 60,0 wt% H₂SO₄) = 28,52 L

Thus, feed ratio (liters 20%  solution/liter 60% solution):

87,8L/28,52L = 3,08

I hope it helps!

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2