Question

How many grams of barium sulfate can be produced from the reaction of 2.54 grams sodium sulfate and 2.54 g barium chloride? Na2SO4(aq) + BaCl2(aq) --> BaSO4(s) + 2NaCl(aq) Report your answer to 3 decimal places.

Answer 1

Answer: 2.796 grams

Step-by-step explanation:

`Na_2SO_4+BaCl_2\rightarrow 2NaCl+BaSO_4`

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

`\text{Number of moles of sodium sulphate}=(2.54g)/(142g/mol)=0.018moles`

`\text{Number of moles of barium chloride}=(2.54g)/(208g/mol)=0.012moles`

According to stoichiometry:

1 mole of
`BaCl_2` reacts with 1 mole of
`Na_2SO_4`

0.012 moles of
`BaCl_2` will react with=
`(1)/(1)* 0.012=0.012moles` of
`Na_2SO_4`

Thus
`BaCl_2` is the limiting reagent as it limits the formation of product.
`Na_2SO_4` is the excess reagent as (0.018-0.012)=0.006 moles are left unused.

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

0.012 moles of
`BaCl_2` will produce=
`(1)/(1)* 0.012=0.012moles` of
`BaSO_4`

Mass of
`BaSO_4=moles* {\text {Molar mass}}=0.012* 233=2.796g`

Thus 2.796 grams of
`BaSO_4` are produced.