How many grams of barium chloride are produced when 1.25x10^23 molecules of HCl reacts with barium according to the following equation?

Answer 1

The balanced equation for the reaction should be- Ba + 2 HCl → BaCl₂ + H₂

We know, 1 mole of a substance contains particles equivalent to Avogadro's number.

So, if 6.023 x 10²³ molecules in 1 mol of HCl

1.25 x 10²³ molecules in (1 * 1.25 x 10²³) / (6.023 x 10²³) = 0.2075 moles of HCL

mole ratio- HCl : BaCl₂ = 2 : 1

So, number of moles of BaCl₂ (n)= 0.2075 / 2 = 0.10375 mol

Molar mass of BaCl₂ = 137.33 + 2*(35.453) = 208.24 g

We know, mass (m) = number of moles (n) * Molar mass (M)

So, mass of BaCl₂ = 0.10375 mol * 208.24 g = 21.6 g