Question

Magnesium metal melts at 625 degree C and boils at 1450 degree C. It's density is 2.85 grams/cc. Assuming an ideal gas calculate the volume occupied by 20.0 grams of magnesium vapor at its boiling point and under a pressure of 1234 torr. Then, if 40.5 grams of magnesium were dissolved in an excess of concentrated HCl, what volume of hydrogen gas would be released at STP and collected over Mercury?

Answer 1

PV= nRT

P = Pressure in atm

V= Volume in Liter

n = no of moles

R = 0.0821 L atm K-1 Mol-1

T = Temperature in Kelvin

Moles of Magnsium = 20 / 24.3 = 0.823 Moles

P = 1234 / 760 = 1.62 atm

V = ? n = 0.823 Moles T = 273 + 1450 = 1723 K

V = nRT / P = 0.823 x 0.0821 x 1723 / 1.62 = 71.86 Liter

Volume of Magnesium = 71.86 Liter

Question 2

Balanced equation:
Mg + 2 HCl ===> MgCl₂ + H₂
Reaction type: single replacement

40.5 grams of magnesium = 40.5 / 24.305 = 1.666 Moles

Moles of H2 produced = 1.6663 Moles

using the above formula we can calculate the volume

P = 1 atm V = ? n = 1.6663 Moles T = 273 K

V = nRT / P = 1.6663 x 0.0821 x 273 / 1 = 37.34 Liter

Volume of H2 gas = 37.34 Liter