__Na(g) + __C(g) > __NaCl(g)

a. Balance the above reaction.
b. What volume of chlorine gas, at STP, is necessary for the complete reaction
of 4.81 g of sodium metal?

Could you please give a step by step explanation? It would be greatly appreciated.

2.35 L

Explanation:

Given data:

Mass of sodium metal = 4.81 g

Volume of Cl react at STP= ?

Balance chemical equation = ?

Solution:

Balance chemical equation:

2Na  +  Cl₂   →  2NaCl

Number of moles of Na:

Number of moles = mass/ molar mass

Number of moles = 4.81 g/ 23 g/mol

Number of moles = 0.21 mol

Now we will compare the moles of sodium and chlorine from balance chemical equation.

                   Na             :          Cl₂

                    2              :             1

                  0.21           :       1/2×0.21 = 0.105 mol

Volume of Cl₂:

PV = nRT

standard temperature = 273 K

standard pressure = 1 atm

R = general gas constant = 0.0821 atm.L/mol.K

n = moles = 0.105 mol

Now we will put the values.

V = 0.105 mol × 0.0821 atm.L/mol.K × 273 K / 1 atm

V = 2.35 L/ 1

V =  2.35 L

i think that the answer is c