For the reaction: Mg + 2 HCl => MgCl2 + H2 a) How many grams of HCl are required to react with 62.0 g of Mg?
To determine how many grams of HCl are required to react with 62.0 g of Mg, we need to use stoichiometry.
1. Start by writing down the balanced chemical equation:
Mg + 2 HCl => MgCl2 + H2
2. Identify the molar mass of Mg and HCl:
The molar mass of Mg = 24.31 g/mol
The molar mass of HCl = 36.46 g/mol
3. Convert the given mass of Mg to moles:
moles of Mg = given mass / molar mass = 62.0 g / 24.31 g/mol = 2.55 mol
4. Use the stoichiometry of the balanced equation to determine the moles of HCl needed:
From the balanced equation, 1 mol of Mg reacts with 2 mol of HCl.
So, moles of HCl = moles of Mg x (2 mol HCl / 1 mol Mg) = 2.55 mol x (2 mol HCl / 1 mol Mg) = 5.10 mol
5. Finally, calculate the mass of HCl needed:
mass of HCl = moles of HCl x molar mass of HCl = 5.10 mol x 36.46 g/mol = 186.0 g
Therefore, 186.0 grams of HCl are required to react with 62.0 grams of Mg.
62.0 g of Mg is 2.55 moles
So, you will need twice that many moles of HCl.
How many grams is that?