You determined that Cr2O3 is the limiting reactant in the reaction below. If the reaction begins with85.00 grams of Cr2O3 and157
1 answer:
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Answer:
The answer is:
(a)
(b) NaCl
(c) 0.211 g
Explanation:
Given:
The mass of NaCl,
= 0.0860 g
The molar mass of NaCl,
= 58.44 g/mol
The volume of ,
= 30.0 ml
or,
= 0.030 L
Molarity of ,
= 0.050 M
Moles of NaCl will be:
=
=
=
now,
Moles of will be:
(a)
The reaction is:
⇒
(b)
1 mole of NaCl react with,
= 1 mol of
0.0015 mol needs,
=
Available mol of NaCl < needed amount of NaCl
So,
The limiting reagent is "NaCl".
(c)
The precipitate formed,
=
=