A 40.8 mol sample of methane gas, CH4(g), is placed in a
1020 L container at 298 K.
What is the pressure, in atm, of the methane gas?
Express the answer using 3 significant figures.
atm

P = 0.979 atm

Explanation:

Given data:

Number of moles of methane = 40.8 mol

Volume of container = 1020 L

Temperature = 298 K

Pressure of gas =?

Solution:

According to ideal gas equation,

PV = nRT

P = pressure

V = volume

n = number of moles

R = general gas constant

T = temperature

Now we will put the values,

P = nRT/V

P = 40.8 mol × 0.0821 atm.L/mol.K × 298 K / 1020 L

P = 998.2 atm /1020

P = 0.979 atm

not clear

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