Question

Calculate the experimental molar volume (L/mol) for an ideal gas at STP using the information that follows. A 15.0 mg piece of solid magnesium was reacted completely with hydrochloric acid in a reaction flask with a volume of 135 mL. The temperature of the reaction was 22.3 °C and the pressure of the gas produced by the reaction was 11.3 kPa. Calculate the volume of hydrogen gas that would have formed in this reaction had it been conducted under standard temperature and pressure conditions (use the combined gas law). Use the volume you have just determined, along with the number of moles of hydrogen gas that would have formed from 15.0 mg of magnesium reactant, to calculate the molar volume of this gas at STP

Answer 1

Volume of hydrogen gas formed under given condition is 135.94 L.

Volume of hydrogen gas formed at STP will be 14.03 liters.

Step-by-step explanation:

`Mg+2HCl\rightarrow MgCl_2+H_2`

Moles of magnesium =
`(15.0 g)/(24 g/mol)=0.625 mol`

According to reaction, 1 mole of magnesium solid gives 1 mol of hydrogen gas.

Then 0.625 mol of magnesium solid will give =:

`(1)/(1)* 0.625 mol=0.625 mol` of hydrogen gas.

Moles of hydrogen gas formed = n = 0.625 moles

Temperature of the gas = T = 22.3 °C =295.45 K

Pressure of the gas = P = 11.3 kPa = 0.11152 atm

(1 atm = 101.325 kPa )

Volume of the gas

`PV=nRT` (ideal gas equation)

`V=(nRT)/(P)`

`V=(0.625 mol* 0.0821 atm l/mol K* 295.45 K)/(0.11152 atm)`

V = 135.94 L

Volume of hydrogen gas formed under given condition is 135.94 L.

Volume of hydrogen gas that would have formed in this reaction had it been conducted under standard temperature and pressure conditions.

`P_1= 0.11152 atm , T_1= 295.45 K, V_1=135.94 L`

`P_2= 1 atm , T_2= 273.15 K, V_2=?`

Using the combined gas equation :

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

`V_2=(0.11152 atm* 135.94 L* 273.15 K)/(295.45 K* 1 atm)`

`V_2=14.03 L`

Volume of hydrogen gas formed at STP will be 14.03 liters.