Question

How many moles of BaSO4 are formed if 0.5 moles of Na2 SO4 react with 60 g of BaCl2? Na2 SO4 + BaCl2 → BaSO4 + 2NaCl

Answer 1

Answer: 0.29

Explanation:

`Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl`

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\textMolar mass}}`

Putting values in above equation, we get:

`\text{Moles of}BaCl_2}=(60g)/(208g/mol)=0.29mol`

Moles of
`Na_2SO_4` = 0.5 moles

According to stochiometry, 1 mole of
`BaCl_2` reacts with 1 mole of
`Na_2SO_4`

Thus 0.29 moles of
`BaCl_2` will react with 0.29 moles of
`Na_2SO_4`

Thus
`BaCl_2` is the limiting reagent as it limits the formation of products and
`Na_2SO_4` is the excess reagent.

1 mole of
`BaCl_2` produces = 1 mole of
`BaSO_4`

0.29 moles of
`BaCl_2` will produce= 0.29 moles of
`BaSO_4`

Answer 2

the balanced equation for the reaction is as follows
Na₂SO₄ + BaCl₂ ----> 2NaCl + BaSO₄
stoichiometry of Na₂SO₄ to BaCl₂ is 1:1
first we need to find out which the limiting reactant is
limiting reactant is fully used up in the reaction.

number of Na2So4 moles - 0.5 mol number of BaCl2 moles - 60 g / 208 g/mol = 0.288 mol
since molar ratio is 1:1 equal number of moles of both reactants should react with each other

therefore BaCl2 is the limiting reactant and Na2SO4 is in excess. amount of product formed depends on number of limiting reactant present.
stoichiometry of BaCl₂ to BaSO₄ is 1:1.

therefore number of BaSO4 moles formed - 0.288 mol