Zinc reacts with aqueous sulfuric acid to form hydrogen gas:

Question

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
.

Answer 1

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.