Question

A student reacted 10.2 g of barium chloride with excess silver nitrate, according to the equation BaCl2(aq)+2AgNO3(aq) +2AgCl(s)+Ba(NO3)2(aq). a. What is the theoretical yield for silver chloride? b. If the student recovers 8.48 g of silver chloride, what is the percent yield for this reaction? Potassium and chlorine combine to form potassium chloride. a. How many grams of potassium chloride can be produced from 2.50 g of potassium? 2k + Cl2 → 2kci mmik= 39.10 ginol o KC1 = 74.55g/mol

Answer 1

BaCl2(aq) + 2AgNO3(aq) ------------> Ba(NO3)2(aq) + 2AgCl(s)

1 mole of BaCl2 react with excess of AgNO3 to gives 2 moles of AgCl

208.23g of BaCl2 react with excess of AgNO3 to gives 2*143.32g of AgCl

10.2g of BaCl2 react with excess of AgNO3 to gives = 2*143.32*10.2/208.23   = 14.04g of AgCl

Theoretical yield of AgCl = 14.04g

b actual yield of AgCl = 8.48g

percent yield = actual yield *100/theoretical yield

                       = 8.48*100/14.04   = 60.4g

a. 2K + Cl2 -------------> 2KCl

2 moles of KCl produced from 2 moles of K

2 moles of K react with excess of Cl2 to gives 2 moles of KCl

2*39g of K react with excess of Cl2 to gives 2*74.5513g of KCl

2.5g of K react with excess of Cl2 to gives = 2*74.5513*2.5/(2*39) = 4.8g of KCl >>>>answer