$Fe+H_{2}SO_{4}\rightarrow FeSO_{4}+H_{2}$$0,01$ $0,01$ $0,01$ $0,01$ (mol)$n_{Fe}=\frac{0,56}{56}=0,01$ (mol)a) $m_{FeSO_{4}}=0,01.152=1,52$ (g)b) $m_{ddH_{2}SO_{4}}=\frac{0,01.98.100}{4,9}=20$ (g)$m_{ddsau}=0,56+20-2.0,01=20,54$ (g)$C$%$_{FeSO_{4}}$$=\frac{1,52}{20,54}.100%=7,4$%Chúc bạn học tốt! ^^
$Fe+H_{2}SO_{4}\rightarrow FeSO_{4}+H_{2}$$0,01$ $0,01$ $0,01$ $0,01$ (mol)$n_{Fe}=\frac{0,56}{56}=0,01$ (mol)a) $m_{FeSO_{4}}=0,01.152=1,52$ (g)b) $m_{ddH_{2}SO_{4}}=\frac{0,01.98.100}{4,9}=20$ (g)$m_{ddsau}=0,56+20-2.0,01=20,54$ (g)$C$%$_{FeSO_{4}}$$=\frac{1,52}{20,54}.100=7,4$%Chúc bạn học tốt! ^^
$Fe+H_{2}SO_{4}\rightarrow FeSO_{4}+H_{2}$$0,01$ $0,01$ $0,01$ $0,01$ (mol)$n_{Fe}=\frac{0,56}{56}=0,01$ (mol)a) $m_{FeSO_{4}}=0,01.152=1,52$ (g)b) $m_{ddH_{2}SO_{4}}=\frac{0,01.98.100}{4,9}=20$ (g)$m_{ddsau}=0,56+20-2.0,01=20,54$ (g)$C$%$_{FeSO_{4}}$$=\frac{1,52}{20,54}.100
%=7,4$%Chúc bạn học tốt! ^^