Zinc reacts with aqueous sulfuric acid to form hydrogen gas:
Zn(s)
+ H2SO4(aq)
→
ZnSO4(aq)
+ H2(g)
In an experiment, 125 mL of wet H2
is collected over water at 27 °C and a barometric pressure of 765torr.
How many grams of Zn
have been consumed? The vapor pressure of water at 27 °C is 26.74 torr
.
To find the number of moles of hydrogen gas produced, we can use the ideal gas law:
PV = nRT
Where:
P = total pressure - vapor pressure of water
V = volume of hydrogen gas collected (125 mL = 0.125 L)
n = number of moles of gas
R = ideal gas constant (0.0821 L atm/mol K)
T = temperature in Kelvin (27°C = 300 K)
Rearranging the formula for n, we get:
n = (P - vapor pressure) * V / RT
n = (765 torr - 26.74 torr) * 0.125 L / (0.0821 L atm/mol K * 300 K)
n = (738.26 torr) * 0.125 L / 24.63
n = 3.75 moles of H2
From the balanced chemical equation, we know that 1 mole of Zn reacts to produce 1 mole of H2. Therefore, the number of moles of Zn consumed is also 3.75 moles.
The molar mass of Zn is 65.38 g/mol. Therefore, the mass of Zn consumed is:
Mass = moles * molar mass
Mass = 3.75 moles * 65.38 g/mol
Mass = 245.925 g
Therefore, approximately 245.93 grams of Zn have been consumed in the reaction.