If 4.644.64 mol of NO2NO2 occupy 25.925.9 L, how many liters will 2.732.73 mol of NO2NO2 occupy at the same temperature and pressure

I will assume that the moles are all mistyped and will change them to:

4.64, 25.9, and 2.73.

15.2 liters

Explanation:

Both situations can be described using the ideal gas law:  

PV=nRT,

where P, V, and T are the pressure, volume, and temperature (in degrees K).  n is the number of moles.  R is the gas constant.

First NO2 :      P1(25.9L) = (4.64moles)RT

Second NO2: P2V2 = (2.73moles)RT2

Divide one equation by the other:

(P1(25.9L)  =  (4.64)RT1)      

  P2V2 = (2.73)RT2          

The temperatures, pressures, and gas constants are unchanged, so we can cancel those to obtain:

(25.9L)/V2 = (4.64/2.73)

V2 = (25.9L)(2.73/4.64)

V2 = 15.2 Liters

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g