analiticka

Submitted by Milenaa on Fri, 05/22/2015 - 18:51
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Sadrzaj hlora u nekom uzorku iznosi 8%. Koliko g uzorka treba uzeti za analizu da bi dobili najmanje 0.1000g taloga srebro hlorida?

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JOVAN - Fri, 05/22/2015 - 22:15

AgNO3 + NaCl ------------------> AgCl + HNO3

 

1molNaCl----------------------------->1molAgCl                                 M(NaCl)= 58.5g/mol

Xg------------------------------------------>0.1000g                                 M(AgCl)=143.5 g/mol

 

58.5g---------------------->143.5g

Xg------------------------>0.1000g

X x 143.5g = 58.5g x 0.1000g

X= 58.5 g x 0.1000g / 143.5g

X=5.85g / 143.5

X=0.04g

ms = mu x m% / 100%

ms= 0.04g x 8% / 100%

ms= 0.0032g NaCl

mozda ovako treba