Question

magnesium metal reacts with aqueous hydrochloric acid to form hydrogen gas and aqueous magnesium chloride. Mg(s)+2HCl(aq)=MgCl2(aq)+H2(g)

A student performs this reaction in a laboratory and collects the hydrogen gas over water. The student collects 215.8mL of gas at a pressure of 755.2mmHg and a temperature of 22 degrees C.

(here are some answers I found already correct me if I am wrong)

I found the water pressure at 22 degrees C to be 19.8mmHg

I found the partial pressure of the hydrogen in the tube to be 735.4mmHg or 0.968atm

Using the ideal gas law to calculate moles of hydrogen produced in the reaction I got 0.0086 mol H2.

I need help with these questions.

How many moles of Mg reacted? (assume 100% yield)

What mass of Mg reacted?

Answer 1

Answer –We are given, reaction - Mg(s)+2HCl(aq) ------> MgCl₂(aq)+H₂(g)

Volume of H₂ gas = 215.8 mL , pressure = 755.2 mm Hg , T = 22 + 273 = 295 K

We know water vapour pressure at 22 oC is 19.83 mm Hg

So vapour pressure of H₂ = 755.2 mm Hg – 19.83 mm Hg

                                        = 735.37 mm Hg

                                        = 0.968 atm

Now using the ideal gas law –

PV=nRT

So, n = PV/RT

         = 0.968 atm * 0.2158 L / 0.0821 L.atm.mol^-1.K^-1 * 295 K

          = 0.00862 moles

So you calculated all up to correctly.

We are given like 100 % yield, means the moles of Mg are totally consumed and formed H₂ gas

Now form the balanced reaction

1 moles of H₂ = 1 moles of Mg

0.00862 moles of H₂ = ?

= 0.00862 moles of Mg

Mass of Mg = 0.00862 moles * 24.305 g/mol

                    = 0.209 g of Mg