molarity question

Question

after a student placed 20.0g of chromiu in 200ml of a H2SO4 solution she collected 2.8 L of H₂ gas at 96.5 kPa and 25 C(Celsius). what was the concentration of the H₂SO₄  solution if she reported a 78% yield for the experiment? the products of the reaction are chromium(iii) sulfatr and hydrogen gas.

 

ive written down the balanced equation and tried many formulas but i still get 0.43 as the molarity? and the answer key says the right answer is 0.70M. help please 

Answer 1

Caroline,

From the phrasing of the question, we see that the goal of the reaction was the collection of H₂ gas. So if 2.8 L was a 78% yield, then we can calculate the theoretical yield as 2.8/.78 = 3.6 L of H₂ gas.

 

The balanced equations for this reaction,  2Cr + 3H₂SO₄ → Cr₂(SO₄)₃ + 3H₂ , serves only to give us the mole ratio between products and reactants. Specifically, for every 3 moles of acid in, we get 3 moles of gas out, or 1:1 ratio. So first we turn our volume, 3.6 L, into a number of moles using pv = nRt.

 

pressure = 96.5 kPa = .952 atm

temp = 25 C = 298 K

 

n = pv/Rt = (.952)(3.6) / (.0821)(298) = 0.14 moles of acid

 

Since molarity = moles/liters  .14moles/0.2L = .7M solution

 

Hope this makes it clear.