Question

How many dm3 of hydrogen are released when 3 g of potassium is reacted with hydroiodic acid?

Answer 1

Answer:

V = 0.0859dm³

Explanation:

Hydroiodic acid, HI, reacts with potassium, K, to produce potassium iodide, KI, and hydrogen, as follows:

2HI + 2K → 2KI + H₂(g)

To solve this question we have to find the moles of hydrogen produced knowing that 2 moles of K produce 1 mole of H₂. With the moles of hydrogen we can find the volume of hydrogen assuming there are STP conditions:

Moles K -Molar mass: 39.0983g/mol-

3g * (1mol / 39.0983g) = 0.0767 moles of K

Moles H₂:

0.0767 moles of K * (1mol H₂ / 2mol K) = 0.03836 moles H₂

Using: PV = nRT; V = nRT / P

Where V is volume in dm³,

n are moles of gas: 0.03836 moles,

R is gas constant = 0.082atm*dm³/molK

T is absolute temperature = 273.15K at STP

and P is pressure = 1atm

The volume of the gas is:

V = 0.03836mol*0.082atm*dm³/molK*273.15K / 1atm

V = 0.0859dm³