Question

what volume of 0.3682 M H2SO4 solution is required to react with 0.4198g of Al(CN)3 according to the reaction, 2Al(CN)3 + 3H2SO4 --> 6HCN + Al2(SO4)3? the answer is 16.28mL, but I don't know how to solve this, please help!!

Answer 1

Have you done other stoichiometry problems before?

Answer 2

no this is the first time, i am taking it for summer

Answer 3

hm okay. We'll the goal is to first convert your reactants into moles.

For this problem, you need to know the molar mass of \(Al(CN)_3\).

Then, to find the moles, use: \(moles=\dfrac{mass}{molar~mass}\)

I'll wait till you do that to explain the rest.

Answer 4

ok I got 105.01 for the molar mass

Answer 5

okay, now find the moles.

Answer 6

so I wrote down 0.004 moles, 0.4198g/105.01g/mol

Answer 7

okay. now, we want to know how many moles of \(H_2SO_4\) we need.

To do so, we need a ratio of the moles of each reactant and their stoichiometric coefficients (from the balanced reaction). n is the symbol for moles

\(\dfrac{n_{H2SO_4}}{3}=\dfrac{n_{Al(CN_3)}}{2}\)

Plug in your values (moles of aluminium cyanide) and solve for moles of sulfuric acid:

\(\dfrac{n_{H2SO_4}}{3}=\dfrac{0.004}{2}\rightarrow n_{H2SO_4}=\dfrac{3*0.004}{2}=0.006 ~moles\)

good so far?

Answer 8

yes

Answer 9

okay, so lastly, they want the answer in terms of volume, given a concentration in Molarity.

So we use the Molarity (M) formula: \(M=\dfrac{n_{solute}}{L_{solution}}\)

Answer 10

Solve for volume (which is in liters here)

Answer 11

yay i got the answer!! thank you so much, i finally understood it! :D

Answer 12

no problem! glad i could help

(just remember that in stoichiometry you need to convert what you're given to moles, then work from there)

Answer 13

yes sir!