Question

When excess nitric acid reacts with 1.68 g of iron, 842 mL of hydrogen gas is collected over water. If the barometric pressure is 833 mmHg and the temperature of the water is 34°C, what is the calculated value of R, in atmL/molK?
(Water vapor pressure at 34°C = 39.90 mmHg)

__Fe(s) + __HNO3(aq) → __Fe(NO3)2 + __H2(g)

Answer 1

Fe(s) + 2HNO3(aq) → Fe(NO3)2 + H2(g)

Molar mass(g/mol) 55.8

Number of moles of Fe reacted , n = mass/molar mass = 1.68 g / (55.8(g/mol)) = 0.030 moles

From the balanced reaction ,

1 mole of Fe produces 1 mole of H2 gas

0.030 moles of Fe produces 0.030 moles of H2 gas

We know that ideal gas equation is PV = nRT

Where

T = Temperature = 34oC = 34 + 273 = 307 K

P = pressure of H2 gas =  barometric pressure - vapor pressure of water

= 833 - 39.90 = 793.1 mm Hg

= 793.1 mm Hg *( 1 atm/ 760 mm Hg)

= 1.043 atm

n = No . of moles = 0.030 moles

R = gas constant = ?

V= Volume of the gas = 842 mL = 0.842 L

Plug the values we get R = (PV) / (nT) = 0.095 L atm/(mol-K)