Question

Nickel metal reacts with sulfuric acid in a redox reaction. In an experiment, 1.25g of nickel reacted with 55.0ml od 0.110M sulfuric acid. Hydrogen gas was collected in a 150.0ml container. The reaction takes place at 20 degrees C. One product of the reaction contained nickel (IV) ion.

a. When the reaction is complete, what will be the pressure of the gas in the container?

b. How many grams of the excess reactant will remain?

Answer 1

Solution :-

Balanced reaction equation

Ni(s) + 2H2SO4 ----- > Ni(SO4)2(aq) + 2 H2(g)

now lets calculate the moles of the Ni and H2SO4

moles of Ni = 1.25 g / 58.6934 g per mol

                     = 0.0213 mol Ni

moles of H2SO4 = 0.110 mol per L * 0.055 L = 0.00605 mol H2SO4

now lets calculate the moles of Ni needed to react with 0.00605 mol H2SO4

0.00605 mol H2SO4 * 1 mol Ni / 2 mol H2SO4 = 0.003025 mol Ni

so the excess moles of Ni remain = 0.0213 mol - 0.003025 mol = 0.018275 mol Ni

So the mass of the excess Ni remain = 0.018275 mol * 58.6934 g per mol = 1.07 g Ni

Now lets calculate the pressure of the gas

lets first calculate the moles of H2 gas that can be produced using the mole ratio of the H2SO4

0.00605 mol H2SO4 * 2 mol H2 / 2 mol H2SO4 = 0.00605 mol H2

volume of container = 150 ml = 0.150 L

temperature T = 20.0 C +273 = 293 K

pressure of gas = ?

PV= nRT

P= nRT/V

P= 0.00605 mol * 0.08206 L atm per mol K * 293 K / 0.150 L

P = 0.970 atm

Therefore the pressure of the gas in the containeer = 0.970 atm

so answers are

a) pressure of gas after reaction = 0.970 atm

b) grams of excess reactant remain = 1.07 g Ni