Bài 1:
a,
$Fe+2HCl\to FeCl_2+H_2$
b,
$n_{H_2}=\dfrac{10,08}{22,4}=0,45(mol)$
$\to n_{Fe}=n_{H_2}=0,45(mol)$
$\to m_{Fe}=0,45.56=25,2g$
c,
$n_{HCl}=2n_{H_2}=0,45.2=0,9(mol)$
$\to C_{M_{HCl}}=\dfrac{0,9}{0,15}=6M$
Bài 2:
a,
$CO_2+Ba(OH)_2\to BaCO_3+H_2O$
b,
$n_{CO_2}=\dfrac{6,72}{2,24}=0,3(mol)$
$\to n_{Ba(OH)_2}=n_{BaCO_3}=n_{CO_2}=0,3(mol)$
$\to C_{M_{Ba(OH)_2}}=\dfrac{0,3}{0,6}=0,5M$
c,
$m_{BaCO_3}=0.3.197=59,1g$