Respuesta :
Answer:
0.676 grams of manganese (IV) oxide should be added.
Explanation:
Moles of chlorine gas = n
Volume of the chlorine gas = V = 205 mL = 0.205 L
Pressure of the chlorine gas = 705 Torr = [tex]\frac{705}{760}atm=0.928 atm[/tex]
1 atm = 760 Torr
Temperature of the chlorine gas = T = 25°C = 25 + 273 K = 298 K
[tex]PV=nRT[/tex] ( ideal gs equation)
[tex]n=\frac{PV}{RT}=\frac{0.928 atm\times 0.205 L}{0.0821 atm L/mol K\times 298 K}=0.00777 mol[/tex]
[tex]MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+2H_2O(l)+Cl_2(g)[/tex]
According to reaction, 1 mole of chlorine gas is obtained from 1 mole of manganese(IV) oxide,then 0.00777 moles of chlorine gas will be obtained from :
[tex]\frac{1}{1}\times 0.00777 mol=0.00777 mol[/tex] of manganese (IV) oxide
Mass of 0.00777 moles of manganese (IV) oxide:
0.00777 mol × 87 g/mol = 0.676 g
0.676 grams of manganese (IV) oxide should be added.
