Question

ce cantitate de oxid de fier(III)reactioneaza cu 227,3 aluminiu impur cu o puritate de 95%? cat fier topit se obtine?

Answer 1

puritatea= masa pura/masa impura*100 => masa pura=95*227,3/100=215,935g Al pur

MAl=27g/mol

n=m/M=215,935/27=8 moli Al

2Al + Fe2O3 => Al2O3 + 2Fe

2 moli Al....................1 mol Fe2O3
8 moli Al...................x mli Fe2O3

x=8/2=

2 moli Al.........................2 moli Fe
8 moli Al........................x=8 moli Fe

MFe=56g/mol

n=m/M=>m=n*M=8*56=