Question

for questions 1-4, show all calculations with proper unit analysis and significant figures. (step by step work please!)

1) what theoretical yield (grams) of hydrogen can be obtained from 0.0334 Mg and 15.00mL of 6.0 M HCL?

2) If the atmosperic pressure is 0.993atm and the height of the water column at the end of the experiement is 16.5cm, what is the total pressure of the gas (mmHg) in the eudiometer?

3)If the total pressure of the gas in the eudiometer at 20.5 degree celcius is 580.4mmHg, what is the partial pressure of the hydrogen gas (mmHg)?

4)If the partial pressure of the hydrogen gas is 513.2mmHg, the volume of gas is 15.4mL, and the temperature of the water is 20.5 degrees celcius, what was the actual yield of hydrogen gas (grams)?

Answer 1

1.

Mg (s) + 2 HCl (aq.) ------------> MgCl2 (aq.) + H2 (g)

Moles of Mg = mass / molar mass = 0.0334 / 24.0 = 0.00139 mol

Moles of HCl = Molarity * Volume / 1000 = 6.0 * 15.00 / 1000 = 0.090 mol

From the balanced equation,

1 mol of Mg required 2 mol HCl

So, 0.00139 mol of Mg requires 2 * 0.00139 = 0.00278 mol of HCl

But we have 0.090 mol of HCl, Hence HCl is excess reagent and Mg is limiting reagent.

From the balanced equation,

1 mol og Mg fomrs 2 mol of H2

Then, 0.00139 mol of Mg forms 2 * 0.00139 mol of H2 = 0.00278 mol of H2

Mass of H2 btained = moles * molar mass = 0.00278 * 2.00 = 0.00556 g.