4 HCl + MnO2 ==> MnCl2 + 2 H2O + Cl2

How many grams of MnCl2 are formed when 36.0 grams of water are also formed?
DO we have to calculate mole mass first?

Yes, that's the only way to find the number of mols in 36.0 g H2O then to convert mols MnCl2 to g MnCl2.

Yes, in order to determine the amount of MnCl2 formed when 36.0 grams of water are also formed, we need to calculate the molar mass of water.

The molar mass of water (H2O) can be calculated by summing the molar masses of its components, which are hydrogen (H) and oxygen (O). The molar mass of hydrogen is approximately 1.008 grams/mol, and the molar mass of oxygen is approximately 16.00 grams/mol.

Therefore, the molar mass of water is calculated as follows:

(2 * molar mass of hydrogen) + molar mass of oxygen
= (2 * 1.008 g/mol) + 16.00 g/mol
= 2.016 g/mol + 16.00 g/mol
= 18.016 g/mol

Now that we have calculated the molar mass of water, which is 18.016 grams/mol, we can use it to determine the amount of MnCl2 formed.

From the balanced chemical equation:
4 HCl + MnO2 ==> MnCl2 + 2 H2O + Cl2

We can see that for every 2 moles of water formed, 1 mole of MnCl2 is produced. This means the molar ratio between water and MnCl2 is 2:1.

To calculate the moles of water from the given mass, we divide the mass of water by its molar mass:

moles of water = mass of water / molar mass of water
= 36.0 g / 18.016 g/mol
≈ 1.998 mol

Since the molar ratio of MnCl2 to water is 1:2, the moles of MnCl2 formed would be half of the moles of water:

moles of MnCl2 = 1.998 mol / 2
= 0.999 mol

Finally, to calculate the mass of MnCl2 formed, we multiply the moles of MnCl2 by its molar mass:

mass of MnCl2 = moles of MnCl2 * molar mass of MnCl2
= 0.999 mol * molar mass of MnCl2

To find the molar mass of MnCl2, we look up the atomic masses of manganese (Mn) and chlorine (Cl) from the periodic table.

The atomic mass of Mn is approximately 54.938 g/mol, and the atomic mass of Cl is approximately 35.453 g/mol.

The molar mass of MnCl2 is calculated as follows:

molar mass of MnCl2 = molar mass of Mn + (2 * molar mass of Cl)
= 54.938 g/mol + (2 * 35.453 g/mol)
= 125.844 g/mol

Now, we can calculate the mass of MnCl2:

mass of MnCl2 = 0.999 mol * 125.844 g/mol
≈ 125.875 g

Therefore, approximately 125.875 grams of MnCl2 are formed when 36.0 grams of water are also formed in this chemical reaction.