[tex] \frac{CuO}{300} [/tex] + [tex] \frac{H _{2}SO _{4} }{x} [/tex] ---> [tex] H_{2} O[/tex] + [tex]CuSO _{4} [/tex]
[tex] M_{H _{2}SO _{4} } = 2A _{H} + A _{S} + 4A _{O} = 2 +32 + 64 = 98 grame H2SO4[/tex]
[tex]M _{CuO} = A _{Cu} + A _{O} = 64 + 16 = 80 grame CuO[/tex]
[tex] \frac{64}{300} = \frac{98}{x} =\ \textgreater \ x = \frac{300 X 98}{64} =\ \textgreater \ x = \frac{29400}{64} =\ \textgreater \ x = 459,375[/tex]
Deci masa sarii obtinute este 459,375
Nu stiu daca este 100% corect, dar sper ca te-am ajutat :)
