A 0.286g sample of Zn was used to produce H2(g)by the reaction. Zn(s)+ 2HCl(aq)>

Question

A 0.286g sample of Zn was used to produce H2(g)by the reaction.

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

The barometric pressure was recorded as 761.6mm Hg and room temperature was 20degreesC. When the water levels of the eudiometer tube and graduated cylinder were equal, the volume of H2 was measured as 94.8mL. Calculate the number of moles of hydrogen gas produced in the reaction. Also, calculate the theoretical number of moles that can be obtained with the given amount of metal. (assume the acid is in excess) Finally, determine the error

Explanation / Answer

m = 0.286 g of Zn

mol = mass/MW = 0.286/65.8

mol of Zn = 0.00437

now.. expect

0.00437 mol of H2 due to 1:1 ratio (theoretical yield)

now..

real yield:

PV = nRT

n = PV/(RT)

Pgas = Ptota - Pvapor = 761.6 -17.5 = 744.1

n = (744.1)(94.8*10^-3)/(62.4 * 293)

n = 0.0038582 mol of H2

c)

yield % = real / theoretical * 100% = 0.0038582/0.00437*100= 88.28%

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