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Răspuns:
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miu,NH3= 17g/mol miu HCl= 36,5g/mol miu NH4CL=53,5g/mol
c= mdx100/ms
a.
calculez masa amoniac din solutie
md= cxms/100= 80x250/100g=200g NH3 m,apa= ms-md= 50g
calculez mada de acid care intra in reactie
17g........36,5g..............53,5g
NH3 + HCL--------. NH4CL
200g...........x.................y
x= m= 429,4g HCl este masa dizolvata in solutia de acid
calculez masa solutiei de acid
ms= mdx100/c= 429,4x100/50g=858,8g solutie
m,apa= ms-md= 858,8g-429,4g=429,4g
b,
solutia finala contine sarea NH4Cl dizolavata in apa care a ramas de la solutiile de NH3 si HCl
calculez masa de NH4Cl ,y din ecuatie
y=200x53,5/17g=629,4g NH4CL
deci md= 629,4g
m,apa= 50g+429,4g=479,4g
m,s= md+m,apa= 629,4+479,4=> 1108,8g
c= mdx100/ms= 629,4x100/1108,8=56,7%
Explicație: