Question

How many moles of silver chloride, produced from 100 g of silver nitrate reacting with barium chloride BaCl2?

Answer 1

Equation of Reaction

2AgNO3 + BaCl2 === 2AgCl + Ba(NO3)2

Molar Mass of AgNO3 = 170g/mol

Moles of reacting AgNO3 = 100g/170gmol-¹

=0.588moles of AgNO3

From the equation of reaction...2moles of AgNO3 reacts to Produce 2Moles of Silver Chloride

So Their ratio is 2:2.

This means that 0.588Moles of AgCl Will be produced too.

ANSWER...0.588MOLES OF AgCl WILL BE PRODUCED.