Đáp án:
Giải thích các bước giải:
a) $2Al+3H2SO4-->Al2(SO4)3+3H2$
x-----------1.5x--------------------------------------1.5x(mol)
$Fe+H2SO4--->FeSO4+H2$
y------y---------------------------------------------->y(mol)
Ta có
$nH2=5,6/22,4=0,25(mol)$
Theo PT
$nH2SO4=nH2=0,25(mol)$
=>$C_M{H2SO4}=0,25/0,4=0,625(M)$
b)
Ta có hpt
$27x+56y=8,3$
$1,5x+y=0,25$
=>$x=0,1,y=0,1$
=>%$mAl=0,1.27/8,3.100$%$=32,53$%
=>%$mFe=100-32,35=67,47$%
c)
$2Al+6H2SO4-->Al2(SO4)3+3SO2+6H2O$
0,1------------------------------------------------------>0,15(mol)$
$2Fe+6H2SO4-->Fe2(SO4)3+3SO2+6H2O$
0,1------------------------------------------------------------>0,15(mol)
=>$nSO2=0,3(mol)$
=>$VSO2=0,3.22,4=6,72(l)$