Can anyone help me with this Chemistry question.

Question

0.080 mg of Mg reacts with HCl (aq) to give 85.5 mL of og H₂ gas at 22^oC and 759 mm Hg pressure.

Mg(s)  + 2HCl (aq)  MgCl₂ (aq) + H₂(g)

a) What is the partial pressure of H₂ gas?

b) Use CGL to calculate volume of H₂ gas at STP.

(hint: make a table with correct units for lab conditions and STP conditions and then solve for V₂).

c) Calculate moles of Mg and get moles of H2 gas using mol-mol ratio from the balanced equation.

d) Calculate molar volume of H₂ gas (Hint: divide V₂ by moles of H₂).

e) Calculate % error (theoretical value = 22.4 L/mol)

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Answer 1

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

0.08 mg...........................................85.5 ml.......@22ºC and 759 mm Hg (assume HCl is in excess)

a) 0.08 mg Mg x 1 mmol Mg/24 mg x 1 mmol H₂/mmol Mg = 3.3x10^-6 moles H₂

Pressure H₂: PV = nRT and P = nRT/V = (3.3x10^-6)(62.36 L-mmHg/Kmol)(295K) / 0.0855 L = 0.710 mm Hg

b) I have no idea what CGL means. But to find volume of H₂ at STP, we can use the idea that 1 mol of an ideal gas @ STP occupies 22.4 L. Therefore..

3.3x10^-6 moles H₂ x 22.4 L/mol = 7.4x10^-5 L of H2 gas at STP

c) moles Mg = 0.08 mg x 1 g/1000 mg x 1mol/24 g = 3.3x10^-6 moles (just as in (a) above)

d) this is probably related to the calculations using CGL and since I have no idea what that is, the calculation will come out the same as theoretical using 22.4 L/mole. If you explain what CGL is, maybe I (or someone) can answer this more effectively.