what volume of hydrogen is produced with excess HCL according to the equation Zn+2HCL-ZnCL2+H2
from the equation 65g of Zn produced 22.4dm3 of H2 :36.5g of Zn=36.5g*22.4dm3/65=12.6dm3
The volume of H2 depends upon how much Zn you started with. You don't have that in the problem.
To determine the volume of hydrogen gas produced when zinc reacts with excess hydrochloric acid (HCl), we need to consider the stoichiometry of the reaction and the ideal gas law.
The stoichiometry of the balanced equation is as follows:
Zn + 2HCl -> ZnCl2 + H2
According to the equation, one mole of zinc (Zn) reacts with two moles of hydrochloric acid (2HCl) to produce one mole of hydrogen gas (H2).
First, we need to determine the amount of zinc present. Let's assume we have 0.1 moles of zinc (Zn). Since the balanced equation indicates a 1:1 ratio between zinc and hydrogen, 0.1 moles of zinc will produce 0.1 moles of hydrogen.
Next, we will use the ideal gas law equation to calculate the volume of hydrogen gas produced. The ideal gas law equation is as follows:
PV = nRT
Where:
P: Pressure (in pascals or atmospheres)
V: Volume (in liters)
n: Number of moles
R: Ideal gas constant (0.0821 L⋅atm/mol⋅K)
T: Temperature (in Kelvin)
Assuming pressure and temperature are constant, we will rearrange the formula to solve for volume (V):
V = (nRT) / P
Let's assume the pressure is 1 atmosphere and the temperature is 273 Kelvin (0 degrees Celsius).
Using the formula, we can calculate the volume of hydrogen gas:
V = (0.1 mol * 0.0821 L⋅atm/mol⋅K * 273 K) / 1 atm
V ≈ 2.24 L
Therefore, approximately 2.24 liters of hydrogen gas will be produced when 0.1 moles of zinc reacts with excess hydrochloric acid.