Câu 9.
a/ PTHH: $Fe+2HCl\to FeCl_2+H_2$
b/ $n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{4,2}{56}=0,075(mol)$
Theo pt: $n_{FeCl_2}=0,075(mol)$
$\to m_{FeCl_2}=n_{FeCl_2}.M_{FeCl_2}=0,075.127=9,525(g)$
c/ Theo pt: $n_{H_2}=0,075(mol)$
$\to V_{H_2}=n_{H_2}.22,4=0,075.22,4=1,68(l)$
Bài 10.
a/ Xét nguyên tố $FeCl_2$
$\%Fe=\dfrac{M_{Fe}.100}{M_{FeCl_2}}\%=44\%$
$\%Cl=\dfrac{M_{Cl}.2.100}{M_{FeCl_2}}\%=56\%$
Xét nguyên tố $HCl$
$\%H=\dfrac{M_H.100}{M_{HCl}}\%=3\%$
$\%Cl=\dfrac{M_{Cl}.100}{M_{HCl}}\%=97\%$
b/ Xét nguyên tố $NaOH$:
$\%Na=\dfrac{M_{Na}.100}{M_{NaOH}}\%=57,5\%$
$\%O=\dfrac{M_O.100}{M_{NaOH}}\%=40\%$
$\%H=\dfrac{M_H.100}{M_{NaOH}}\%=2,5\%$
Xét nguyên tố $KNO_3$:
$\%K=\dfrac{M_K.100}{M_{KNO_3}}\%=38\%$
$\%N=\dfrac{M_N.100}{M_{KNO_3}}\%=14\%$
$\%O=\dfrac{M_O.3.100}{M_{KNO_3}}\%=48\%$
