Las disoluciones comerciales de HCl (PF = 36.5 g/mol) típicamente son 39.0% P/P de este ácido en agua. Determine la molalidad de
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Answer:
Are produced 12,1 g of AlCl₃ and 5,33 g of Al₂O₃ are left over
Explanation:
For the reaction:
Al₂O₃ (s) + 6 HCl (aq) → 2AlCl₃ (aq) + 3H₂0(l)
10,0g of Al₂O₃ are:
10,0g ₓ = <em>0,0980 moles</em>
And 10,0g of HCl are:
10,0 gₓ = <em>0,274 moles</em>
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For a total reaction of 0,274 moles of HCl you need:
0,274× = <em>0,0457 moles of Al₂O₃</em>
Thus, limiting reactant is HCl
The grams produced of AlCl₃ are:
0,274 moles HCl × × 133
= <em>12,1 g of AlCl₃</em>
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The moles of Al₂O₃ that don't react are:
0,0980 moles - 0,0457 moles =<em> </em>0,0523 moles
And its mass is:
0,0523 molesₓ = <em>5,33 g of Al₂O₃ </em>
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I hope it helps!
Answer:
"1 M" will be the right solution.
Explanation:
The given values are:
Number of moles,
= 2 moles
Volume of solution,
= 2000 mL
or,
= 2 L
Now,
The molarity of the solution will be:
=
On substituting the values, we get
=
=
Answer : Two electrons balances the charges in this half reaction.
Explanation :
The given half reaction is,
In this half reaction, there is no charges present on the left side of the reaction but on right side of the reaction, (2+) charge is present. So, two electrons (2-) are added on the right side of the reaction for balances the charge of the half reaction.
The balanced half reaction is,
Hence, two electrons balances the charges in this half reaction.