Answer to Question #120107 in Inorganic Chemistry for ariel jacob

Question

one of the methods utilized in the preparation of potassium nitrate(KNO3) is the reaction between nitric acid (HNO3) and potassium carbonate(K2CO3). for this experiment, small amount of solid potassium carbonate are added gradually to 5mL of 30% w/v nitric acid until effervescence is no longer observed. 1.775 grams of purified KN03 were obtained. what is the percentage yield for this reaction? show all steps involved in your calculations.

Answer 1

Actual yield of KNO3 =1.775g

Mass of HNO3= 5ml of 30% w/v

30% means 0.3 mole/L; therefore there's (5/1000)*0.3 =0.0015 moles in 5 ml of HNO3.=0.0015*64. =0.096 g

$Yield =\frac{theoretical}{actual} *100$

Yield = (0.096/1.775)* 100 =5.4%