When 25.0 g of Zn reacts, how many L of gas are formed at and a pressure of 854 mmHg?
A)22.4 L
B)8.32 L
C)0.120 L
D)0.382 L
E)8.56 L

Question

When  25.0 g of  Zn reacts, how many L of  gas are formed at  and a pressure of  854 mmHg?
A)22.4 L 
B)8.32 L 
C)0.120 L 
D)0.382 L 
E)8.56 L

Quizplus · Accepted Answer

The balanced equation for the reaction between Zn and HCl is:

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

Using the stoichiometry of the equation, we can find the amount of H2 produced from 25.0 g of Zn:

25.0 g Zn × (1 mol Zn/65.38 g Zn) × (1 mol H2/1 mol Zn) × (22.4 L H2/1 mol H2) = 8.32 L H2

Note that we used the molar volume of a gas at STP (22.4 L/mol) to convert from moles of H2 to liters of H2.

The pressure and temperature given in the question are not at standard conditions, but we can use the ideal gas law to adjust for this:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

We can rearrange the equation to solve for the volume of gas at the given conditions:

V = nRT/P

Plugging in the values we know:

n = 1 mol (from the stoichiometry of the reaction)
R = 0.0821 L·atm/(mol·K) (the ideal gas constant)
T = 298 K (25°C converted to Kelvin)
P = 854 mmHg (converted to atm by dividing by 760 mmHg/atm)

V = (1 mol)(0.0821 L·atm/mol·K)(298 K)/(0.854 atm) = 2.91 L

Therefore, the answer is B) 8.32 L, since this is the amount of H2 produced from 25.0 g of Zn.

When 250 G of Zn Reacts, How Many L of L