Al 0.792 g excess H2SO4 H2 gas 0.0813 g percent

Al 0.792 g excess H2SO4 H2 gas 0.0813 g percent

 

 

Aluminum metal reacts with sulfuric acid to form hydrogen gas and aluminum sulfate.

a) Write a balanced chemical equation for this reaction.

b) Suppose that a 0.792 g sample of aluminum that contains impurities is reacted with excess sulfuric acid and 0.0813 g of H2 is collected. Assuming that none of the impurities reacts with sulfuric acid to produce hydrogen, what is the percentage of aluminum in the sample?

 

 

balanced chemical equation

2Al + 3H2SO4 → 3H2 + Al2(SO4)3

 

 

 

H2의 몰질량 = 2.0159 g/mol 이므로,

0.0813 g / (2.0159 g/mol) = 0.04033 mol H2

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

2Al + 3H2SO4 → 3H2 + Al2(SO4)3

 

Al : H2 = 2 : 3 계수비(= 몰수비) 이므로,

Al : H2 = 2 : 3 = ? mol : 0.04033 mol

 

? = 2 × 0.04033 / 3 = 0.02689 mol Al

 

 

 

Al의 몰질량 = 26.9815 g/mol 이므로,

0.02689 mol × (26.9815 g/mol) = 0.7255 g Al

---> 반응한 Al의 질량

= Al 시료 중 순수한 Al의 질량

 

 

 

Al 시료의 순도를 계산하면,

0.7255 / 0.792 × 100

= 91.6%

 

 

 

불순물의 함량을 계산하면,

100 – 91.6

= 8.4%

 

또는

(0.792 – 0.7255) / 0.792 × 100

= 8.396%

 

 

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H2의 몰질량 = 2.02 g/mol 이므로,

0.0813 g / (2.02 g/mol) = 0.0402 mol H2

 

 

Al : H2 = 2 : 3 계수비(= 몰수비) 이므로,

Al : H2 = 2 : 3 = ? mol : 0.0402 mol

 

? = 2 × 0.0402 / 3 = 0.0268 mol Al

 

 

Al의 몰질량 = 26.98 g/mol 이므로,

0.0268 mol × (26.98 g/mol) = 0.723 g Al

 

 

Al 시료의 순도를 계산하면,

0.723 / 0.792 × 100 = 91.3%

 

 

불순물의 함량을 계산하면,

100 – 91.3 = 8.7%

 

또는

(0.792 – 0.723) / 0.792 × 100 = 8.71%