In a mixture of argon and hydrogen, occupying a volume of 1.18 l at 894.6 mmhg and 44.1oc, it is found that the total mass of the sample is 1.25 g. what is the partial pressure of argon?

To solve the question, there is a need to use the equation:

PV = nRT

(894.6/760) × 1.18 = n × 0.0821 × (273 + 44.1)

By solving we get:

Total moles, n = 0.053

Assume, the moles of argon as a and of hydrogen as b,

So,

40 × a + 2 × b = 1.25 (i)

a + b = 0.053 (ii)

By solving i and ii we get:

a = 0.03,

Thus, mole fraction of Ar = XAr = 0.03/0.053 = 0.57

So,

Partial pressure of Ar = 894.6 × XAr = 894.6 × 0.57 = 509.92 mm Hg

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