\(n_M=a\left(mol\right)\)
\(n_{M_2O}=b\left(mol\right)\)
\(m_{hh}=aM+b\left(2M+16\right)=2.16\left(g\right)\)
\(\Leftrightarrow M\cdot\left(a+b\right)+16b=2.16\left(1\right)\)
\(M+H_2O\rightarrow MOH+\dfrac{1}{2}H_2\)
\(M_2O+H_2O\rightarrow2MOH\)
\(n_{HCl}=0.04\cdot1=0.04\left(mol\right)\)
\(MOH+HCl\rightarrow MCl+H_2O\)
\(0.04.......0.04\)
\(n_{MOH}=a+2b=0.04\left(mol\right)\left(2\right)\left(0\le b\le0.02\right)\)
\(\text{Thay (2) vào (1) : }\)
\(\Leftrightarrow0.04M+16b=2.16\)
\(\Leftrightarrow b=\dfrac{2.16-0.04M}{16}\)
\(\Leftrightarrow0\le\dfrac{2.16-0.04M}{16}\le0.02\)
\(\Leftrightarrow46\le M\le54\)
Sai đề !
Gọi $n_M = a(mol) ; n_{M_2O} = b(mol)$
$\Rightarrow Ma + b(2M + 16) = 2,16 (1)$
$2M + 2H_2O \to 2MOH + H_2$
$M_2O + H_2O \to 2MOH$
$MOH + HCl \to MCl + H_2O$
$n_{MOH} = a + 2b = n_{HCl} = 0,04.1 = 0,04(mol)$
\(Ma + 2Mb + 16b = 2,16\\ \Rightarrow M(a + 2b) + 16b = 2,16\\ \Rightarrow 0,04M + 16b = 2,16\)
Vì 0 < b < 0,02
nên 46 < M < 54
Không có kim loại nào thỏa mãn
