화학량론. 44.23 g Cu + HNO3

화학량론. 44.23 g Cu + HNO3

 

 

A 44.23 g sample of metallic copper is added to 136 mL of 6.0 M nitric acid to produce aqueous copper(II) nitrate, liquid water and gaseous nitrogen monoxide.

a. Write a balanced chemical equation.

b. What is the limiting reagent?

c. How many grams of nitrogen monoxide are produced?

 

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a.

3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2(aq) + 4H2O(l) + 2NO(g)

 

 

b.

Cu의 몰질량 = 63.55 g/mol

44.23 g / (63.55 g/mol) = 0.696 mol Cu

( 계산 설명 http://ywpop.tistory.com/7738 )

 

(6.0 mol/L) * 0.136 L = 0.816 mol HNO3

 

Cu : HNO3 = 3 : 8 = 0.696 : ?

? = 8 * 0.696 / 3 = 1.856 mol HNO3

---> HNO3는 이만큼 없다.

 

Cu : HNO3 = 3 : 8 = ? : 0.816

? = 3 * 0.816 / 8 = 0.306 mol Cu

---> Cu는 남는다. 과잉 반응물.

---> 따라서 HNO3가 한계 반응물.

( 설명 http://ywpop.tistory.com/3318 )

 

 

c.

HNO3 : NO = 8 : 2 = 4 : 1 = 0.816 : ?

? = 0.816 / 4 = 0.204 mol NO

 

NO의 몰질량 = 30.01 g/mol

0.204 mol * (30.01 g/mol) = 6.12 g NO