H2 is produced by the reaction of 118.5 ml of a 0.8775-m solution of h3po4 according to the following equations: 2cr +
2h3po4 ? 3h2 + 2crpo4.
(a) outline the steps neccessary to determine the number of moles and mass of h2
(b)perform the calculations outlined

Explanation:

a) First we will calculate moles of phosphoric acid present in 118.5 mL of a 0.8775 M solution.

After determining the moles of phosphoric acid, we will then calculate the number pf moles of hydrogen gas by the help unitary method.

From moles of hydrogen gas mass of the hydrogen gas can be calculated easily by the formula:

=Moles of compound × molecular mass of compound

b) Moles of phosphoric acid in  in 118.5 mL of a 0.8775 M solution be n

n = 0.10398 mol

According to reaction , 2 moles of phosphoric acid gives 3 moles of hydrogen gas:

Then 0.10398 moles of phosphoric acid will give:

of hydrogen gas

0.15597 moles of hydrogen gas are produced.

Mass 0.15597 moles of hydrogen gas:

0.15597 mol × 2 g/mol=0.31194 g

0.31194 grams of hydrogen gas are produced.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.