Respuesta :
Answer:
0.88 g
Explanation:
Using ideal gas equation to calculate the moles of chlorine gas produced as:-
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 805 Torr
V = Volume of the gas = 235 mL = 0.235 L
T = Temperature of the gas = [tex]25^oC=[25+273]K=298K[/tex]
R = Gas constant = [tex]62.3637\text{torr}mol^{-1}K^{-1}[/tex]
n = number of moles of chlorine gas = ?
Putting values in above equation, we get:
[tex]805torr\times 0.235L=n\times 62.3637\text{ torrHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{805\times 0.235}{62.3637\times 298}=0.01017\ mol[/tex]
According to the reaction:-
[tex]MnO_2+4HCl\rightarrow MnCl_2+2H_2O+Cl_2[/tex]
1 mole of chlorine gas is produced when 1 mole of manganese dioxide undergoes reaction.
So,
0.01017 mole of chlorine gas is produced when 0.01017 mole of manganese dioxide undergoes reaction.
Moles of [tex]MnO_2[/tex] = 0.01017 moles
Molar mass of [tex]MnO_2[/tex] = 86.93685 g/mol
So,
[tex]Mass=Moles\times Molar\ mass[/tex]
Applying values, we get that:-
[tex]Mass=0.01017moles \times 86.93685\ g/mol=0.88\ g[/tex]
0.88 g of [tex]MnO_2(s)[/tex] should be added to excess HCl (aq) to obtain 235 mL of [tex]Cl_2(g)[/tex] at 25 degrees C and 805 Torr.
