0.300 M NaOH 105.0 mL 0.060 M AlCl3 150.0 mL

0.300 M NaOH 105.0 mL 0.060 M AlCl3 150.0 mL

 

 

A solution of 105.0 mL of 0.300 M NaOH is mixed with

a solution of 150.0 mL of 0.060 M AICI3.

a) Write the balanced chemical equation for the reaction that occurs.

b) What precipitate forms?

c) What is the limiting reactant?

d) How many grams of this precipitate form?

e) What is the concentration of each ion that remains in solution?

 

 

 

a) Write the balanced chemical equation for the reaction that occurs.

 

AlCl3(aq) + 3NaOH(aq) → Al(OH)3(s) + 3NaCl(aq) 또는

3NaOH(aq) + AlCl3(aq) → Al(OH)3(s) + 3NaCl(aq)

 

AlCl3 + 3NaOH → Al(OH)3 + 3NaCl 또는

3NaOH + AlCl3 → Al(OH)3 + 3NaCl

 

 

 

b) What precipitate forms?

 

Al(OH)3

 

 

 

c) What is the limiting reactant?

 

각 반응물의 몰수를 계산하면,

> (0.300 M) (105.0/1000 L) = 0.0315 mol NaOH

( 참고 https://ywpop.tistory.com/7787 )

 

> (0.060) (150.0/1000) = 0.0090 mol AlCl3

 

 

 

3NaOH(aq) + AlCl3(aq) → Al(OH)3(s) + 3NaCl(aq)

 

NaOH : AlCl3 = 3 : 1 계수비(= 몰수비) 이므로,

0.0315 mol NaOH와 반응하는 AlCl3의 몰수를 계산하면,

NaOH : AlCl3 = 3 : 1 = 0.0315 mol : ? mol

 

? = 1 × 0.0315 / 3 = 0.0105 mol AlCl3

---> AlCl3는 이만큼 없다, 부족하다?

---> AlCl3 = 한계 반응물

( NaOH = 과잉 반응물 )

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

 

 

300x250

 

 

d) How many grams of this precipitate form?

 

생성물의 양은 한계 반응물의 양에 의해 결정되고,

AlCl3 : Al(OH)3 = 1 : 1 계수비(= 몰수비) 이므로,

침전되는 Al(OH)3의 몰수 = 0.0090 mol

 

 

 

Al(OH)3의 몰질량 = 78.00 g/mol 이므로,

0.0090 mol × (78.00 g/mol) = 0.702 g Al(OH)3

( 참고 https://ywpop.tistory.com/7738 )

 

---> 0.70 g

 

 

 

e) What is the concentration of each ion that remains in solution?

 

전체 부피 = 105.0 + 150.0 = 255.0 mL

= 0.2550 L

 

 

 

NaOH : AlCl3 = 3 : 1 = ? mol : 0.0090 mol

 

? = 3 × 0.0090 = 0.027 mol NaOH

---> 0.0090 mol AlCl3와 반응한 NaOH의 몰수

 

 

 

0.0315 – 0.027 = 0.0045 mol NaOH

( 유효숫자 처리 생략 )

---> 반응하고 남은 NaOH의 몰수

= OH^-의 몰수

 

 

 

[Na^+] = 0.0315 mol / 0.2550 L

= 0.124 M

( Na^+는 침전되는 양이 없다. )

 

 

 

[OH^-] = 0.0045 mol / 0.2550 L

= 0.0176 M

---> 0.018 M

 

 

 

0.0090 mol AlCl3 로부터,

3 × (0.0090 mol) = 0.027 mol Cl^-

 

 

 

[Cl^-] = 0.027 mol / 0.2550 L

= 0.106 M

---> 0.11 M

 

 

 

 

[키워드] AlCl3 + NaOH 반응 기준, NaOH + AlCl3 반응 기준