a sample of methane gas (ch4) with a mass of 58 g is kept in a 1.5 L container at a temperature of 373 K. what is the pressure of the gas?

Given :

A sample of methane gas (CH₄) with a mass of 58 g is kept in a 1.5 L container at a temperature of 373 K.

To Find :

The pressure of the gas.

Solution :

Moles of methane :

We know, by ideal gas equation :

PV = nRT

Here, R is ideal gas equation,

So,

Hence, this is the required solution.

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