Mg(OH)2 yield 88.2% Mg mass 7.73 g water 1.31 g H2 mass

Mg(OH)2 yield 88.2% Mg mass 7.73 g water 1.31 g H2 mass

 

 

If the reaction yield is 88.2%,

what mass in grams of hydrogen is produced

by the reaction of 7.73 g of magnesium with 1.31 g of water?

Mg(s) + 2H2O(l) → Mg(OH)2(s) + H2(g)

 

---------------------------------------------------

 

각 반응물의 몰수를 계산하면,

7.73 g / (24.305 g/mol) = 0.318 mol Mg

( 참고 https://ywpop.tistory.com/7738 )

 

 

1.31 g / (18.015 g/mol) = 0.0727 mol H2O

 

 

 

Mg + 2H2O → Mg(OH)2 + H2

 

0.318 mol Mg와 반응하는 H2O의 몰수를 계산하면,

Mg : H2O = 1 : 2 = 0.318 mol : ? mol

 

? = 2 × 0.318 = 0.636 mol H2O

---> H2O는 이만큼 없다, 부족하다?

---> H2O = 한계 반응물.

( 참고: 한계 반응물 https://ywpop.tistory.com/3318 )

 

 

 

H2O : H2 = 2 : 1 = 0.0727 mol : ? mol

 

? = 0.0727 / 2 = 0.03635 mol H2

 

 

 

0.03635 mol × (2.016 g/mol) = 0.07328 g H2

---> 이론적 수득량

 

 

 

수율을 고려하면,

0.07328 g × (88.2/100) = 0.0646 g H2

 

 

 

답: 0.0646 g

 

 

 

 

[키워드] Mg(OH)2 yield 88.2% Mg 7.73 g water 1.31 g H2 gas mass

 

[FAQ] [①23/03/30]