heo me ai làm hộ mình vs ạ đang gấppp vclllllll

$\rm CTDC:R_2(CO_3)_3\\0,4863M_{R_2(CO_3)_3}=M_{Na_2SO_4}\\\xrightarrow[]{}M_{R_2(CO_3)_3}=142:0,4863=292(amu)\\\xrightarrow[]{}2M_R+(12+16.3).3=292\\\xrightarrow[]{}M_{R}=56(amu)\\\xrightarrow[]{}R:Fe(Iron)$

Đáp án:

Ta có: $\rm \dfrac{M_{Na_2SO_4}}{M_B}=0,4863\to M_B=292\ (g/mol)$

Gọi muối có công thức: $\rm R_2(CO_3)_3$.

$\rm M_{R_2(CO_3)_3}=2M_R+(12+16×3)×3=292\ (g/mol)$

$\rm\to M_R=\dfrac{292-180}{2}=56\ (g/mol)$

Tên: Iron, ký hiệu: Fe.