Chemistry Question?

Question

I was wondering if someone could help me with this problem.
The question is: Calculate the volume of 0.17M sulfuric acid that would be needed to neutralize 194 mL of a 0.44 M aqueous ammonia solution. The equation for the reaction is:

H2SO4 (aq) + 2 NH3 (aq) ------- 2 NH4 (aq) + SO4 -2 (aq)

Answer 1

Remember that volume x concentration = mass. Also remember the ratios you get from the chemical equation. In this case, 1 mol of H2SO4 is required to react with 2 mols NH3. The calculation sets up like:

(0.44 M NH3)(0.194 L)(1 mol H2SO4/2 mol NH3)/(0.17 M H2SO4)

Answer 2

194 mls. of 0.44M NH3 contains 0.194x 0.44=0.08536 M of NH3. So, according to the equation which you give, you need half this number of moles of H2SO4 to neutralise it. That's 0.04268 moles. So if the strength of your H2SO4 solution is 0.17M/L, you'll need 0.04268/0.17 litres, which is 0.2511 litres, which is 251.1 mls.

Answer 3

0.44M means 0.44 mole in 1000 ml
i.e. 0.08536 moles in 194 ml

1 H2SO4 for 2 NH3

So 1/2(0.08536)= 0.04268 moles

0.17 moles in 1000 ml of H2SO4

0.04268 in .........251.058 ml.