If 0.33 moles of sulfuric acid were consumed in the reaction, how many
moles of ammonium hydroxide would be required?

Sulfurous acid is formed by the reaction of water and sulfur dioxide.

Balanced reaction will provide the basic information.

Balanced Reaction :

                         SO₂ +  H₂O > H₂SO₃

                        1 mol    1mol                  1mol

From above reaction we come to know

1 mole of H₂O ≅ 1 mole H₂SO₃

if 1 mole of H₂O produces 1 mole of H₂SO₃ then

0.680 mole of water will be required to produces sulfurous acid

Explanation through calculation

apply the unity formula

1 mole of the H₂O ≅ 1 mole of H₂SO₃

x mole of the H₂O ≅  0.680 moles of H₂SO₃

so by cross multiplication

1 mole  x X mole ≅ 1 mole  x 0.680 mole

we have to find moles of H₂O so by rearrangement of the above equation

X mole of H₂O = 1 mole x 0.680 mole

X mole of H₂O =  0.680 moles

Explanation:

K = 3.3

Explanation:

Nitric acid, HNO3, reacts with nitrogen monoxide, NO, to produce nitrogen dioxide, NO2 and water H2O as follows:

2HNO3(g) + NO(g) → 3NO2(g) + H2O(g)

Where equilibrium constant, K, is:

K = [NO2]³[H2O] / [HNO3]²[NO]

[] is the molar concentration of each species at equilibrium.

To solve this question we need to find molarity of each gas and replace these in the equation as follows:

[NO2] -Molar mass NO2-46.0g/mol-

18.6g * (1mol/46.0g) = 0.404mol / 7.7L = 0.0525M

[H2O] -Molar mass:18.01g/mol-

236.7g * (1mol/18.01g) = 13.14 moles / 7.7L = 1.707M

[HNO3] -Molar mass:53.01g/mol-

16.2g * (1mol/53.01g) = 0.3056 moles / 7.7L = 0.0397M

[NO] -Molar mass: 30.0g/mol-

11.0g * (1mol/30.0g) = 0.367 moles / 7.7L = 0.0476M

Replacing:

K = [NO2]³[H2O] / [HNO3]²[NO]

K = [0.0525M]³[1.707M] / [0.0397M]²[0.0476M]