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Bibi2071
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Peste 5 g Zn se adauga o solutie de HCl de c=36,5%. Ce volum de solutie de HCl trebuie introdusa in eprubeta stiind ca densitatea solutiei de HCl e de c=36,5 este de 1,133k/m cubi ??
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Răspuns :

Moonlight01
Zn + 2HCl => ZnCl2 + H2

65g Zn.....................2*36,5g HCl
5g Zn.........................xg HCl

x=73*5/65=5,615g HCl (md)

C=md/ms*100=>ms=md*100/C=5,615*100/36,5=15,3846g sol. HCl

ro=ms/V => V=ms/ro=15,3846*10^(-3)/1,133kg/m3 = 13,5786*10^(-3)m3 = 13,5786L HCl

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