Question

A quality control analyst mixes 30.0mL of 0.255M of BaCl2 with excess AgNo3 causing AgCl to precipipate and subsequently isolated 2.01g of AgCl. show a balance equation for the reaction and calculate the percent yield of the reaction​

Answer 1

Given info : A quality control analyst mixes 30.0mL of 0.255M of BaCl2 with excess AgNo3 causing AgCl to precipipate and subsequently isolated 2.01g of AgCl.

To find : the percentage yield of the reaction is ..

solution : first we should write balance chemical equation.

BaCl₂ + 2AgNO₃ ⇒ 2AgCl + Ba(NO₃)₂ , this is the balanced chemical equation of given reaction.

moles of BaCl₂ = molarity * volume in L = 0.255 * 30/1000 = 0.00765 mol

we see that one mole of barium chloride gives two moles of silver chloride.

so, 0.0765 mol of barium chloride will give 2 * 0.0765 = 0.153 mol

so the mass of 0.153 mol of silver chloride = no of moles of silver chloride * molecular weight of silver chloride

                                                                       = 0.0153 mol * 143.32 g/mol

                                                                       = 2.192796 ≈ 2.2g

now percentage yield = experimental yield/theoretical yield * 100

                                     = 2.1/2.2 * 100 = 95.45 %

therefore the percentage yield of the reaction is 95.45 %