25.0 cm3 NaOH(aq), 2.24 mol/dm3, 3.86 g crystals

25.0 cm3 NaOH(aq), 2.24 mol/dm3, 3.86 g crystals

 

 

Using 25.0 cm3 of aqueous sodium hydroxide, 2.24 mol/dm3, 3.86 g of crystals were obtained. Calculate the percentage yield.

2NaOH + H2SO4 → Na2SO4 + 2H2O

Na2SO4 + 10H2O → Na2SO4•10H2O

 

Number of moles of NaOH used =

Maximum number of moles of Na2SO4•10H2O that could be formed =

Mass of one mole of Na2SO4•10H2O = 322 g

Maximum yield of sodium sulphate-10-water = g

Percentage yield = %

 

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25.0 cm3 = 25.0 mL

2.24 mol/dm3 = 2.24 mol/L = 2.24 M

( 설명 https://ywpop.tistory.com/9330 )

 

 

2.24 M NaOH 용액 25.0 mL에 들어있는 NaOH의 몰수를 계산하면,

(2.24 mol/L) × 0.0250 L = 0.0560 mol NaOH

Number of moles of NaOH used =

 

 

NaOH : Na2SO4 = 2 : 1 계수비(= 몰수비) 이므로,

생성물 Na2SO4의 몰수를 계산하면,

NaOH : Na2SO4 = 2 : 1 = 0.0560 mol : ? mol

? = 0.0560 / 2 = 0.0280 mol Na2SO4

= 0.0280 mol Na2SO4•10H2O

Maximum number of moles of Na2SO4•10H2O that could be formed =

 

 

Na2SO4•10H2O의 몰질량 = 322 g/mol 이므로,

0.0280 mol Na2SO4•10H2O의 질량을 계산하면,

0.0280 mol × (322 g/mol) = 9.02 g Na2SO4•10H2O

Maximum yield of sodium sulphate-10-water = g

 

 

(3.86 g / 9.02 g) × 100 = 42.8%

Percentage yield = %

( 계산 설명 https://ywpop.tistory.com/61 )