2Al+6HCl-to>2AlCl3+3H2
\(\dfrac{8}{15}\)------1,6--------\(\dfrac{8}{15}\)---0,8
n H2=\(\dfrac{17,92}{22,4}\)=0,8 mol
=>m Al=\(\dfrac{8}{15}\).27=14,4g
=>m HCl=1,6.36,5=58,4g
=>m AlCl3=\(\dfrac{8}{15}\).133,5=71,2g
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