Sulphuric acid (H2SO4) solution reacts with magnesium metal (Mg) to
magnesium sulphate and hydrogen according to:
H2SO4 + Mg → MgSO4 + H2
How many litres of 5.00 M sulphuric acid would be required to dissolve
20.00 g of magnesium and how many mol of hydrogen would be
produced?
(RMM data: H = 1.01, S = 32.07, O = 16.00, Mg = 24.31)
mols Mg = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Mg to mols H2SO4.
Then M H2SO4 = mols H2SO4/L H2SO4. You know mols and M, solve for L.
Using mols Mg, convert to mols H2 for part 2.
m,mm
To determine the number of liters of 5.00 M sulfuric acid required to dissolve 20.00 g of magnesium, you need to follow these steps:
1. Calculate the molar mass of magnesium (Mg).
- Mg molar mass = 24.31 g/mol
2. Convert the mass of magnesium to moles.
- Moles of Mg = mass of Mg / molar mass of Mg
- Moles of Mg = 20.00 g / 24.31 g/mol
3. Write a balanced chemical equation for the reaction:
H2SO4 + Mg → MgSO4 + H2
4. Determine the stoichiometry of the reaction.
- From the equation, you can see that 1 mole of Mg reacts with 1 mole of H2SO4. Therefore, the moles of H2SO4 required would be the same as the moles of Mg.
5. Calculate the volume of 5.00 M sulfuric acid required using the molarity formula:
Molarity = moles / volume (in liters)
- Convert the moles of H2SO4 required to liters of solution using the molarity formula:
Volume (in liters) = Moles of H2SO4 / Molarity of H2SO4
- Moles of H2SO4 = Moles of Mg calculated in step 2
- Molarity of H2SO4 = 5.00 mol/L
To determine the number of moles of hydrogen produced, you need to follow these steps:
1. Determine the moles of magnesium used (from step 2).
2. From the balanced chemical equation, you can see that the ratio of H2 to Mg is 1:1.
Therefore, the number of moles of hydrogen produced would be the same as the moles of magnesium used.
Note: Remember to use the appropriate significant figures in your calculations and make sure to include the unit (mol) with your final answer.