Respuesta :
Answer:
Volume of hydrogen gas formed under given condition is 135.94 L.
Volume of hydrogen gas formed at STP will be 14.03 liters.
Explanation:
[tex]Mg+2HCl\rightarrow MgCl_2+H_2[/tex]
Moles of magnesium = [tex]\frac{15.0 g}{24 g/mol}=0.625 mol[/tex]
According to reaction, 1 mole of magnesium solid gives 1 mol of hydrogen gas.
Then 0.625 mol of magnesium solid will give =:
[tex]\frac{1}{1}\times 0.625 mol=0.625 mol[/tex] 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
[tex]PV=nRT[/tex] (ideal gas equation)
[tex]V=\frac{nRT}{P}[/tex]
[tex]V=\frac{0.625 mol\times 0.0821 atm l/mol K\times 295.45 K}{0.11152 atm}[/tex]
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.
[tex]P_1= 0.11152 atm , T_1= 295.45 K, V_1=135.94 L[/tex]
[tex]P_2= 1 atm , T_2= 273.15 K, V_2=?[/tex]
Using the combined gas equation :
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
[tex]V_2=\frac{0.11152 atm\times 135.94 L\times 273.15 K}{295.45 K\times 1 atm}[/tex]
[tex]V_2=14.03 L[/tex]
Volume of hydrogen gas formed at STP will be 14.03 liters.
