Respuesta :
Answer:
Oxide of M is [tex]M_2O_3[/tex] and sulfate of [tex]M_2(SO_4)_3[/tex]
Explanation:
0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.
Let moles of hydrogen gas be n.
Temperature of the gas ,T= 17°C =290 K
Pressure of the gas ,P= 741 mmHg= 0.9633 atm
Volume occupied by gas , V = 0.303 L
Using an ideal gas equation:
[tex]PV=nRT[/tex]
[tex]n=\frac{PV}{RT}=\frac{0.9633 atm\times 0.303 L}{0.0821 atm L/mol K\times 290 K}=0.01225 mol[/tex]
Moles of hydrogen gas produced = 0.01225 mol
[tex]2M+2xHCl\rightarrow 2MCl_x+xH_2[/tex]
Moles of metal =[tex]\frac{0.225 g}{27.0 g/mol}=8.3333 mol[/tex]
So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.
[tex]\frac{8.3333}{0.01225 mol}=\frac{2}{x}[/tex]
x = 2.9 ≈ 3
[tex]2M+6HCl\rightarrow 2MCl_3+3H_2[/tex]
[tex]MCl_3\rightarrow M^{3+}+Cl^-[/tex]
Formulas for the oxide and sulfate of M will be:
Oxide of M is [tex]M_2O_3[/tex] and sulfate of [tex]M_2(SO_4)_3[/tex].
