Answer:
1.368 grams of manganese dioxide must b added to HCl to obtain 385 mL of chlorine gas.
Explanation:
Using ideal gas equation:
PV = nRT
where,
P = Pressure of chlorine gas = ![765 Torr=\frac{765}{760}atm](https://tex.z-dn.net/?f=765%20Torr%3D%5Cfrac%7B765%7D%7B760%7Datm)
1 atm = 760 Torr
V = Volume of chlorine gas = 385 mL = 0.385 L ( 1 mL - 0.001 L)
n = number of moles of chlorine gas = ?
R = Gas constant = 0.0821 L.atm/mol.K
T = Temperature of chlorine gas =25 °C= 25 + 273 K = 300 K
Putting values in above equation, we get:
![(\frac{765 }{760}atm)\times 0.385 L=n\times (0.0821L.atm/mol.K)\times 300K\\\\n=0.01573 mole](https://tex.z-dn.net/?f=%28%5Cfrac%7B765%20%7D%7B760%7Datm%29%5Ctimes%200.385%20L%3Dn%5Ctimes%20%280.0821L.atm%2Fmol.K%29%5Ctimes%20300K%5C%5C%5C%5Cn%3D0.01573%20mole)
![MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+2H_2O(l)+Cl_2(g)](https://tex.z-dn.net/?f=MnO_2%28s%29%2B4HCl%28aq%29%5Crightarrow%20MnCl_2%28aq%29%2B2H_2O%28l%29%2BCl_2%28g%29%20)
According to reaction, 1 mole of chlorine gas is obtained from 1 mole of manganese dioxide ,then 0.01573 moles of chlorine gas will be obtained from :
of manganese dioxide
Mass of 0.01573 moles of manganese dioxide:
0.01573 mol × 86.94 g/mol = 1.368 g
1.368 grams of manganese dioxide must b added to HCl to obtain 385 mL of chlorine gas.
Sorry I don’t exactly know I have exams rn I need points
I think that the answer might be A) 1.5 L of NH3(g)
Silver. It is number 47 on the periodic table meaning it has 47 protons. It’s atomic mass is 108. 47+61=108 so it is definitely silver