Công thức : $m_{}$ =$n_{}$.$M_{}$
$V_{}$ = $n_{}$. $22,4_{}$
$n_{}$ = $C_{M}$. $V_{}$
a,$m_{Fe}$ = $n_{Fe}$.$M_{}$ =$0,1_{}$.$56_{}$ =$5,6_{}$g
b,$n_{CO2}$ = $\frac{V}{22,4}$= $\frac{3,36}{22,4}$= $0,15_{}$mol
$m_{CO2}$=$0,15_{}$.$44_{}$=$6,6_{}$g
c, Đổi 100ml= 0,1(l)
$n_{NaOH}$=$1.0,1_{}$ =$0,1_{}$ mol
$m_{NaOH}$ = $0,1.40_{}$=$4 g_{}$