mdd= 800+100 = 900 (g)
a) C% = \(\dfrac{800}{900}\).100% = 88,89%
b)n\(_{CuSO_4}\)= \(\dfrac{800}{160}\)= 5 (mol)
PT: CuSO4 + 2NaOH ---> Cu(OH)2\(\downarrow\) + Na2SO4
mol:__5------------------------>5
m\(_{Cu\left(OH\right)_2}\)= 5 . 98 = 490 (g)