How many grams of sodium are required to completely react with 19.2 L of Cl, gas at STP according to the following chemical reaction?
Remember 1 mol of an ideal gas has a volume of 22.4 L at STP 2 Na(s) +
Cl2(g) – 2 NaCl(s)

12438 grams of sodium are required to completely react with 19.2 L of Cl, gas at STP.

Explanation:

Data given:

mass of sodium required = ?

volume of Chlorine gas = 19.2 L or 19200 grams

Balanced chemical reaction:

2 Na(s) + (g) ⇒ 2 NaCl(s)

moles of chlorine gas in the reaction is calculated

atomic mass of chlorine gas = 71 grams/litre

number of moles of chlorine gas =

number of moles of chlorine gas =

number of moles = 270.4 moles

from the balanced reaction,

2 moles of Na, and 1 mole of chlorine gas reacts to form 2 moles of NaCl

270.4 moles of chlorine gas is provided so number of moles of Na will be multiplied by 2, using mole ratio

so number of moles of Na = 270.4 x 2

                                            =  540.8 moles

mass of sodium = number of moles x atomic mass

mass of sodium = 540.8 x 23

                            = 12438 grams