how many moles of hydrogen will be liberated from 200.0ml of 3.00M HCL reacting with an excess of magnesium?
Mg + 2HCl = H2 + MgCl2
mols HCl = M x L = ?
mols H2 = 1/2 that (from the coefficients)
To determine the number of moles of hydrogen liberated from the reaction between HCl and magnesium, we need to use the balanced chemical equation for the reaction and the stoichiometry.
The balanced chemical equation for the reaction between HCl and magnesium is:
2 HCl + Mg -> MgCl2 + H2
According to the equation, for every 2 moles of HCl, 1 mole of H2 is produced.
To calculate the number of moles of HCl in 200.0 mL of a 3.00 M solution, we can use the formula:
moles of solute = volume of solution (L) x molarity (mol/L)
First, convert the volume of the solution from mL to L:
200.0 mL = 200.0 mL * (1 L / 1000 mL) = 0.200 L
Then, calculate the moles of HCl:
moles of HCl = 0.200 L * 3.00 mol/L = 0.600 mol
Since the reaction is based on the stoichiometric ratio of 2 moles of HCl to 1 mole of H2, the number of moles of H2 liberated will be half of the moles of HCl:
moles of H2 = 0.600 mol / 2 = 0.300 mol
Therefore, 0.300 moles of hydrogen will be liberated from the reaction.