Question

How many moles of HCI are required to produce 1.50 kg of MnCI2?

Answer 1

To determine the number of moles of HCl required to produce 1.50 kg of MnCl2, first, we need to know the molar mass of MnCl2.

The molar mass of MnCl2 can be calculated by adding the molar masses of manganese (Mn) and chlorine (Cl):

Molar mass of Mn = 54.938 g/mol
Molar mass of Cl = 35.453 g/mol

Molar mass of MnCl2 = (54.938 g/mol) + 2*(35.453 g/mol) = 125.864 g/mol

Next, convert the mass of MnCl2 from kg to grams:

1.50 kg = 1500 g

Now, we can calculate the number of moles of MnCl2 using the formula:

moles = mass (g) / molar mass (g/mol)

moles of MnCl2 = 1500 g / 125.864 g/mol ≈ 11.94 mol

Since the balanced chemical equation for the reaction between HCl and MnCl2 is:

2HCl + MnCl2 -> 2MnCl + H2

We can see that it takes 2 moles of HCl to produce 1 mole of MnCl2. Therefore, to produce 11.94 moles of MnCl2, we need:

11.94 moles MnCl2 * 2 moles HCl / 1 mole MnCl2 ≈ 23.88 moles HCl

So, approximately 23.88 moles of HCl are required to produce 1.50 kg of MnCl2.