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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.

1 Answer

7 votes

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.

User GtEx
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