Question

A quantity control analyst mixes 30.0ml of 0.255M BaCl2 with excess AgNO3 causing AgCl to precipitate and subsequently isolates 2.01g of AgCl. Show a balanced equation for the reaction and calculate the percent yield of the reaction. (BaCl2:208.23g/mol, AgCl:143.32g/mol)

Answer 1

2AgNO₃ (aq) + BaCl₂ (aq) → 2AgCl (s) + Ba(NO₃)₂ (aq)

$Moles = Molarity × Volume(l)$

$Moles(BaCl_2)=0.2555×\frac{30.0}{1000}=0.008moles$

$Moles (AgCl)= 2×0.008=0.016moles$

$Mass(AgCl)=0.016×143.32=1.147grams$