How many grams of zinc metal are required to produce 10.0L of hydrogen gas at 25 degrees C and 1.00 atm pressure, by reaction with excess hydrochloric acid?

Question

Zn (s) + 2HCl (aq) -> ZnCl2 (aq) + H2 (g)

Answer 1

First write the balanced equation for the reaction of HCl with Zn (NOTE: the previous answer did NOT do this and thus obtained an incorrect answer)

Zn + 2HCl ==> ZnCl₂ + H₂(g)

moles of H2 desired: PV = nRT

n = PV/RT = (1atm)(10L)/(0.0821 Latm/Kmol)(298K)

n = 0.409 mole H2 desired

moles Zn needed:

0.409 moles H2 x 1 mole Zn/2 moles HCl = 0.204 moles Zn needed

mass Zn needed = 0.204 moles Zn x 65.4 g/mol = 13.4 g Zn needed (to 3 sig. figs.)

Answer 2

To solve this we are going to use the ideal gas law formula, PV=nRT, where R=0.08206L-atm/mol-K, n is the number of moles of gas, P is the pressure in atm,, T is the temperature in Kelvin and V is the volume in liter. This applies to the hydrogen gas.

PV=nRT ---> n= PV/RT

Let the mass of zinc be m and the molecular weight of zinc be M.

The above reaction implies that the number of moles of Zinc is equal to the number of moles of Hydrogen has. This can be translated as follows:

Number of moles of zinc=(PV/RT). Given that the number of moles=m/M, we get:

(m/M)= (PV/RT)--->m= M x (PV/RT).

We need to convert the temperature to Kelvin. T= 25 degrees C +273= 298K

Upon plugging in the numbers, we get: m= 65.38 g/mol x (1atm x 10.0/0.08206L-atm/mol-K x298K)= 26.7g

Thus, we have shown that 26.7 grams of zinc are required to produce 10.0L of hydrogen gas under the given conditions.