Zn + 2HCl -> ZnCl2 + H2
If 7.29 g of Zinc reacted, how many liters of HCl were used?
I know that i can't use mass volume conversion, because the HCl is not a gas...so what do i do?
Very observant. Obviously the person making up the question meant to include a molarity of HCl but didn't. No way to work the problem.
okay, thank you!
To determine the number of liters of HCl used in the reaction, you need to use stoichiometry and the molar ratio between Zn and HCl.
First, you need to determine the number of moles of Zn used in the reaction. This can be done by dividing the given mass of Zn (7.29 g) by its molar mass (65.38 g/mol for Zn).
Number of moles of Zn = Mass of Zn / Molar mass of Zn
Number of moles of Zn = 7.29 g / 65.38 g/mol
Number of moles of Zn = 0.1114 mol
From the balanced chemical equation, you can see that the molar ratio between Zn and HCl is 1:2. This means that for every 1 mole of Zn, 2 moles of HCl are required.
To find the number of moles of HCl, you need to multiply the number of moles of Zn by the molar ratio:
Number of moles of HCl = Number of moles of Zn * Molar ratio between Zn and HCl
Number of moles of HCl = 0.1114 mol * 2
Number of moles of HCl = 0.2228 mol
Now, you need to convert the moles of HCl into liters. To do this, you can use the ideal gas law equation, assuming that HCl behaves like an ideal gas under the given conditions (standard temperature and pressure):
PV = nRT
Where:
P = Pressure (atm)
V = Volume (liters)
n = Number of moles
R = Gas constant (0.0821 L.atm/mol.K)
T = Temperature (in Kelvin)
Since we are looking for the volume (V), we rearrange the equation:
V = nRT / P
Assuming standard temperature (273 K) and pressure (1 atm), we can substitute the values into the equation:
V = 0.2228 mol * 0.0821 L.atm/mol.K * 273 K / 1 atm
V = 5.873 L
Therefore, approximately 5.873 liters of HCl were used in the reaction.