What mass of barium sulfate can be produced by
complete reaction of 27.4 g of na2so4?

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

MM:                    142.04       233.29

              Ba²⁺ + Na₂SO₄ ⟶ BaSO₄ + 2Na⁺

m/g:                      27.4

(a) Moles of Na₂SO₄

(b) Moles of BaSO₄

(c) Mass of BaSO₄

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