A4.10-mole sample of helium gas at 28°c takes up a volume of 3.28 l. determine the pressure of helium in units of atm.

n = number of mole of sample of helium gas = 4.10

T = temperature of the gas = 28 °C = 28 + 273 = 301 K

V = Volume of helium gas = 3.28 L = 0.00328 m³

P = pressure of the helium gas

Using the equation

P V = n R T

P (0.00328) = (4.10) (8.314) (301)

P = 3.13 x 10⁶ Pa

P = (3.13 x 10⁶ Pa) (1 atm/(101325 Pa))

P =  (3.13 x 10⁶ ) atm/(101325)

P = 30.89 atm

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

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