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Alex777 [14]
3 years ago
14

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

Chemistry
2 answers:
faust18 [17]3 years ago
6 0
<span>the balanced equation for the reaction is as follows
Na</span>₂<span>SO</span>₄<span> + BaCl</span>₂<span> ----> 2NaCl + BaSO</span>₄ 
<span>stoichiometry of Na</span>₂<span>SO</span>₄<span> to BaCl</span>₂<span> 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 BaCl</span>₂<span> to BaSO</span>₄<span> is 1:1.
                                                                           
therefore number of BaSO4 moles formed - 0.288 mol</span>
ozzi3 years ago
6 0

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}=\frac{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

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