100% ........................ 250g Al impur
90% ........................ m.p. = 225 g Al pur
a)
225g md g n moli
2Al + 6HCl --> 2AlCl3 + 3H2
2x27 6x36,5 3
=> md = 225x6x36,5/2x27 = 912,5 g HCl
c% = mdx100/ms
=> ms = mdx100/c% = 912,5x100/36,5 = 250 g sol. HCl
b)
n = 225x3/2x27 = 12,5 moli H2