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Answer:
in 3 mole of
Explanation:
It is clear that in the given of
have
of
Therefore 3 moles of will have
of
Since 1 ion of anything is equivalent to
Therefore 3 moles of will have
of
Which is equivalent to
Thus 3 moles of gives
of
.
Answer:
is the formula for the limiting reagent.
Mass of silver chloride produced is 71.8 g.
Explanation:
Moles of silver nitrate = 0.500 mol
Moles of copper(II) chloride = 0.285 mol
According to reaction, 2 moles of silver nitrate reacts with 1 mole of copper chloride , then 0.500 mole of silver nitrate will react with :
of copper(II) chloride
As we can see that moles of copper(II) chloride will be reacting is 0.250 mol less than present moles of copper (II) chloride ,so this means that silver nitrate is limiting reagent.
And moles of silver chloride to be formed will depend upon silver nitrate.
According to reaction, 2 moles of silver nitrate gives 2 moles of silver chloride , then 0.500 mole of silver nitrate will give :
of silver chloride
Mass of silver chloride produced:
0.500 mol × 143.5 g/mol = 71.8 g