a) The completely balanced chemical reaction is:
Zn(s) + H2SO4(aq)
--------> ZnSO4(aq) + H2 (g)
<span>b) Initially, we determine the quantity of zinc that has reacted based on the produced H2.</span>
According to stoichiometry, 1 mole of Zn is required for each mole of H2 created, thus:
moles(Zn) = moles(H2)
where moles are calculated as the ratio of mass to molar mass (MM)
mass(Zn) / MM(Zn) = mass(H2) / MM(H2)
mass(Zn) = [mass(H2) / MM(H2)] * MM(Zn)
mass(Zn) = [(0.0764 g)/(2 g/mol)] * 65.38 g/mol
mass(Zn) = 2.49 g
Consequently, we find 2.49 g of pure zinc in the sample, leading to a purity of zinc of:
purity = (2.49 / 3.86) * 100 % = 64.50 %
<span>c) In part (b), it is assumed that the impurities in the sample do not react with sulfuric acid to emit hydrogen.
Thus, the hydrogen solely arises from the reaction of Zn with sulfuric acid.</span>
Respuesta:
67.1%
Explicación:
Según la ecuación química, al calcular los moles de carbonato de sodio, se puede determinar la cantidad de NaHCO₃ que reaccionó y su masa, por lo tanto:
Moles Na₂CO₃ - 105.99g/mol-:
6.35g * (1mol / 105.99g) = 0.0599 moles de Na₂CO₃ son producidos.
Como se produce 1 mol de carbonato de sodio al reaccionar 2 moles de NaHCO₃, la cantidad de moles de NaHCO₃ que reaccionaron es:
0.0599 moles de Na₂CO₃ * (2 moles NaHCO₃ / 1 mol Na₂CO₃) = 0.1198 moles de NaHCO₃
Y la masa de NaHCO₃ en la muestra (masa molar: 84g/mol):
0.1198 moles de NaHCO₃ * (84g / mol) = 10.06g de NaHCO₃ estaban en la muestra original.
El porcentaje de NaHCO₃ en la muestra es:
10.06g NaHCO₃ / 15g muestra * 100 =
67.1%
