Mg+2HCl>>H2 + MgCl2
moles of Mg= 2.335/atomicmassMg
so you get the same number of moles of H2.
gramsH2=molesH2*molmassH2
moles of Mg= 2.335/atomicmassMg
so you get the same number of moles of H2.
gramsH2=molesH2*molmassH2
1. Write down the balanced chemical equation for the reaction:
Mg + 2HCl -> MgCl2 + H2
2. Calculate the number of moles of magnesium (Mg):
Given mass of Mg = 2.335 g
Molar mass of Mg = 24.31 g/mol
Moles of Mg = mass of Mg / molar mass of Mg
3. Calculate the limiting reactant:
Since we have the moles of Mg, we need to compare it with the moles of HCl.
Moles of HCl = concentration of HCl x volume of HCl solution
Given concentration of HCl = 4.00 M
Given volume of HCl solution = 50.0 mL = 0.0500 L
Moles of HCl = concentration of HCl x volume of HCl solution
The mole ratio between Mg and HCl in the balanced equation is 1:2.
Compare the moles of Mg with the moles of HCl to determine the limiting reactant.
4. Determine the moles of H2 produced:
Since one mole of magnesium produces one mole of hydrogen gas, the moles of hydrogen gas produced will be the same as the moles of magnesium used.
5. Calculate the mass of H2 gas:
Mass of H2 = Moles of H2 x molar mass of H2
Let's calculate it step by step:
Step 1:
The balanced chemical equation is: Mg + 2HCl -> MgCl2 + H2
Step 2:
Molar mass of Mg = 24.31 g/mol
Moles of Mg = 2.335 g / 24.31 g/mol
Step 3:
Moles of HCl = 4.00 M x 0.0500 L
Step 4:
Since the mole ratio between Mg and HCl is 1:2, we have twice the number of moles of HCl compared to Mg. Therefore, the limiting reactant is Mg. So, the moles of H2 produced will be the same as the moles of Mg used.
Step 5:
Molar mass of H2 = 2 g/mol
Mass of H2 = Moles of H2 x molar mass of H2
Now, you can plug in the values and calculate the mass of H2.