What volume (in liters, at 253 K and 6.54 atm) of chlorine gas is required to react with 7.89 g of P? 2P(s) + 5Cl2(g) → 5PCl5(s)

7.89 grams of phosphorus reacts with 45.15 grams of chlorine

Explanation:

For reacting with 2 mole of P react with 5 mole of Cl2 gas

Molecular weight of phosphorus = 30.97

Molecular weight of 2 mole of phosphorus = 2 * 30.97 = 61.94 grams

Molecular weight of Chlorine molecule = 70.9 grams

Weight of 5 moles of Chlorine molecule = 5 * 70.9 grams = 354.5 grams

61.94 grams of phosphorus reacts with 354.5 grams of chlorine

7.89 grams of phosphorus reacts with 354.5/61.94 * 7.89  grams of chlorine

7.89 grams of phosphorus reacts with 45.15 grams of chlorine

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

the correct answer is:

a. formation of precipitation