200. g Ca3P2 100. g H2O yield PH3 57.4 g

200. g Ca3P2 100. g H2O yield PH3 57.4 g

 

 

200. g Ca3P2와 100. g H2O가 반응할 때,

PH3의 이론적 수득량은 57.4 g이다.

Ca3P2 + 6H2O → 3Ca(OH)2 + 2PH3

 

 

Phosphine, PH3, can be prepared by

the hydrolysis of calcium phosphide, Ca3P2:

Ca3P2 + 6H2O → 3Ca(OH)2 + 2PH3

Based on this equation,

which of the following statements are correct?

Show evidence to support your answer.

a) One mole of Ca3P2 produces 2 mol of PH3.

b) One gram of Ca3P2 produces 2 g of PH3.

c) Three moles of Ca(OH)2 are produced

for each 2 mol of PH3 produced.

d) The mole ratio between phosphine and

calcium phosphide is 2 mol PH3 over 1 mol Ca3P2.

e) When 2.0 mol of Ca3P2 and 3.0 mol of H2O react,

4.0 mol of PH3 can be formed.

f) When 2.0 mol of Ca3P2 and 15.0 mol of H2O react,

6.0 mol of Ca(OH)2 can be formed

g) When 200. g of Ca3P2 and 100. g of H2O react,

Ca3P2 is the limiting reactant.

h) When 200. g of Ca3P2 and 100. g of H2O react,

the theoretical yield of PH3 is 57.4 g.

 

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h) When 200. g of Ca3P2 and 100. g of H2O react,

the theoretical yield of PH3 is 57.4 g.

 

각 반응물의 몰수를 계산하면,

 

Ca3P2의 몰질량 = 182.18 g/mol 이므로,

200. g / (182.18 g/mol) = 1.0978 mol Ca3P2

( 참고 https://ywpop.tistory.com/7738 )

 

H2O의 몰질량 = 18.015 g/mol 이므로,

100. / 18.015 = 5.55 mol H2O

 

 

 

Ca3P2 + 6H2O → 3Ca(OH)2 + 2PH3

 

Ca3P2 : H2O = 1 : 6 계수비(= 몰수비) 이므로,

1.0978 mol Ca3P2와 반응하는 H2O의 몰수를 계산하면,

Ca3P2 : H2O = 1 : 6 = 1.0978 mol : ? mol

 

? = 6 × 1.0978 = 6.5868 mol H2O

---> H2O는 이만큼 없다, 부족하다?

---> H2O = 한계 반응물.

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

 

 

 

생성물의 양은 한계 반응물에 의해 결정되므로,

H2O : PH3 = 6 : 2 = 5.55 mol : ? mol

 

? = 2 × 5.55 / 6 = 1.85 mol PH3

---> 이론적 수득량(mol)

 

 

 

PH3의 몰질량 = 34.00 g/mol 이므로,

1.85 mol × (34.00 g/mol) = 62.9 g PH3

---> 이론적 수득량(g)

 

 

 

답: 62.9 g