Question

80 g roztworu NaOH o stężeniu 20% reaguje z kwasem siarkowym VI . Ile g soli powstaje ?

Answer 1

M(NaOH)=23g+16g+1g=40g

M(Na₂SO₄)=2*23g+32g+4*16g=142g

 

mr=80g

Cp=20%

ms=?

 

Cp=ms/mr * 100%    /*mr

Cp*mr=ms*100%    /:100%

ms=Cp*mr/100%

 

ms=20%*80g/100%

ms=16g

 

2NaOH+H₂SO₄-->Na₂SO₄+2H₂O

16g-------------------xg

80g-------------------142g

 

xg=16g*142g/80g

xg=28,4g