Question

2HNO (ag) +

LMg(s) →
Mg(NO3)2(s)
Helg)
155.0-ml of a solution of 0.75 M HNO3 is combined with 270 g of magnesium. What
volume of hydrogen gas will be produced at STP?

Answer 1

The volume occupied by the gas is 1.302 × 10⁻³ m³

Step-by-step explanation:

2HNO₃(aq) + Mg(s) → Mg(NO₃)₂ + H₂(g)

From the above reaction, we have that 2 moles of HNO₃ combine with 1 mole of Mg to produce 1 mole of Mg(NO₃)₂ and 1 mole of H₂

In the question, we have 155.0 mL of a solution of 0.75 M HNO₃ combining with 270 g of magnesium

Therefore, number of moles of HNO₃ = 0.75×155.0/1000 = 0.11625 moles

`Number \, of \, moles \ of \ magnesium = (Mass \ of \, magnesium)/(Molar \, mass \ of \, magnesius) = (270)/(24.305) = 11.11 \ moles`

Therefore, HNO₃ is the limiting reaction, hence;

0.11625 moles of HNO₃ will react with 1/2×0.11625 moles or 0.058125 moles of Mg to produce 0.058125 moles of H₂

Hence since we now know the volume of H₂ produce, we can find the volume occupied at STP by the following universal gas equation relationship;

PV = nRT

Therefore;

`V = (n * R * T)/(P)`

Where:

V = Volume of occupied by the gas = Required quantity

n = Number of moles = 0.058125 moles

R = Universal gas constant = 8.3145 J/(mol·K)

T = Temperature of the gas at STP = 273.15 K

P = Pressure of the gas at STP = 1 atm = 101325 Pa

`V = (0.058125 * 8.3145 * 273.15)/(101325) = 1.302 * 10^(-3) m^3`