How many liters of H2 gas can be produced at 0 ∘C and 1.00 atm (STP) from 48.0 g of Zn?
Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)
mols Zn = grams/atomic mass = ?
Using the coefficients in the balanced equation, convert mols Zn to mols H2.
Then convert mols H2 to L using mols H2 x 22.4 L/mol = ?L.
To calculate the number of liters of H2 gas produced, we need to use the given balanced chemical equation and the stoichiometry of the reaction.
First, we'll determine the number of moles of Zn present in 48.0 g of Zn. We do this by using the molar mass of Zn, which is 65.38 g/mol:
Number of moles of Zn = Mass of Zn / Molar mass of Zn
Number of moles of Zn = 48.0 g / 65.38 g/mol ≈ 0.7348 mol
From the balanced chemical equation, we can see that the ratio between Zn and H2 is 1:1. So, 1 mole of Zn will produce 1 mole of H2 gas.
Therefore, the number of moles of H2 gas produced is also 0.7348 mol.
Now, we need to use the ideal gas law to calculate the volume of H2 gas at STP (Standard Temperature and Pressure). At STP, temperature (T) is 0 °C or 273.15 K, and pressure (P) is 1.00 atm.
The ideal gas law is given as:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Ideal Gas Constant (0.0821 L•atm/mol•K)
T = Temperature (in Kelvin)
Rearranging the equation to solve for V (volume):
V = (nRT) / P
Substituting the values into the equation:
V = (0.7348 mol) * (0.0821 L•atm/mol•K) * (273.15 K) / (1.00 atm)
V ≈ 16.0 L
Therefore, approximately 16.0 liters of H2 gas can be produced at 0 °C and 1.00 atm from 48.0 g of Zn.