Question

how many grams of silver chloride (AgCI) can be produced if you start with 4.62 grams of barium chloride (BaCI2)

Answer 1

The chemical equation representing the reaction of silver nitrate with barium chloride:

$2AgNO_{3}(aq) + BaCl_{2}(aq)--> 2AgCl (s) + Ba(NO_{3})_{2}(aq)$

Given mass of barium chloride = 4.62 g

Moles of $BaCl_{2} = 4.62 g BaCl_{2}*\frac{1 mol BaCl_{2}}{208.23 g BaCl_{2}} = 0.0222 mol BaCl_{2}$

Moles of AgCl = $0.0222 mol BaCl_{2} * \frac{2 mol AgCl}{1 mol BaCl_{2}}$ = 0.0444 mol AgCl

Mass of AgCl = $0.0444 mol AgCl * \frac{143.32 g AgCl}{1 mol AgCl} = 6.36 g AgCl$

Answer 2

The answer to your question would be 6.27 grams

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