Astudent performed an experiment in the same manner as part b (magnesium+hcl). the temperature for the day was 22°c in the barometric pressure was 742mmhg. 43.5 ml of gas was collected how many grams of magnesium must’ve been used in the reaction?

Mass =  0.048 g

Explanation:

Given data:

Temperature =22 °C (273+ 22 = 296K)

Pressure = 742 mmHg (742/760 = 0.98 atm)

Volume = 43.5 mL ( 0.044 L)

Mass of magnesium = ?

Solution:

PV = nRT

n = PV/RT

n =  0.98 atm × 0.044 L / 0.0821 atm. L. mol⁻¹. K⁻¹ × 296K)

n = 0.043 atm. L / 24.3 atm. L. mol⁻¹

n = 0.002 mol

Mass of Mg:

Mass = number of moles ×  molar mass

Mass =  0.002 mol × 24 g/mol

Mass =  0.048 g