Question

a 40.0 mg sample of a metal reacts with hcl according to the equation in the text. 24.4 ml of gas at 20.0 o c is collected by the downward displacement of aqueous acid as described in the lab protocol. if the atmospheric pressure is 780 mm hg, what is the atomic weight of the metal?

Answer 1

The atomic weight of the metal is 36846.15 g/mol.

Step 1: Calculate the moles of gas produced

n = PV/RT

where:

n is the number of moles of gas

P is the pressure in atm (780 mmHg / 760 mmHg/atm = 1.026 atm)

V is the volume of gas in liters (24.4 mL / 1000 mL/L = 0.0244 L)

R is the ideal gas constant (0.0821 L atm/mol K)

T is the temperature in Kelvin (20.0 °C + 273.15 = 293.15 K)

Plugging in the values, we get:

n = (1.026 atm)(0.0244 L) / (0.0821 L atm/mol K)(293.15 K) = 0.000108 mol

Step 2: Calculate the moles of metal reacted

According to the balanced chemical equation, 1 mole of metal reacts with 1 mole of HCl to produce 1 mole of H₂ gas. Therefore, the number of moles of metal reacted is equal to the number of moles of H₂ gas produced.

n(metal) = n(H₂) = 0.000108 mol

Step 3: Calculate the atomic weight of the metal

atomic weight = mass of metal / moles of metal

atomic weight = (40.0 mg) / (0.000108 mol) = 36846.15 g/mol .