Answer:
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Answer:
Explanation:
This is a limiting reactant problem.
Mg(s)
+
2HCl(aq)
→
MgCl
2
(
aq
)
+ H
2
(
g
)
Determine Moles of Magnesium
Divide the given mass of magnesium by its molar mass (atomic weight on periodic table in g/mol).
4.86
g Mg
×
1
mol Mg
24.3050
g Mg
=
0.200 mol Mg
Determine Moles of 2M Hydrochloric Acid
Convert
100 cm
3
to
100 mL
and then to
0.1 L
.
1 dm
3
=
1 L
Convert
2.00 mol/dm
3
to
2.00 mol/L
Multiply
0.1
L
times
2.00 mol/L
.
100
cm
3
×
1
mL
1
cm
3
×
1
L
1000
mL
=
0.1 L HCl
2.00 mol/dm
3
=
2.00 mol/L
0.1
L
×
2.00
mol
1
L
=
0.200 mol HCl
Multiply the moles of each reactant times the appropriate mole ratio from the balanced equation. Then multiply times the molar mass of hydrogen gas,
2.01588 g/mol
0.200
mol Mg
×
1
mol H
2
1
mol Mg
×
2.01588
g H
2
1
mol H
2
=
0.403 g H
2
0.200
mol HCl
×
1
mol H
2
2
mol HCl
×
2.01588
g H
2
1
mol H
2
=
0.202 g H
2
The limiting reactant is
HCl
, which will produce
0.202 g H
2
under the stated conditions.
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<u>Answer:</u> The molality of the solution is 0.1 m.
<u>Explanation:</u>
To calculate the molality of solution, we use the equation:
Where,
= Given mass of solute = 27.1 g
= Molar mass of solute = 27.1 g/mol
= Mass of solvent = 100 g
Putting values in above equation, we get:
Hence, the molality of the solution is 0.1 m.
Q. How many molecules of H2O can be produced from reactants in container below?
A. 3 molecule of molecules H2O will be produced from reactants in container.
<em><u>Explanation</u><u>:</u></em>
There are seven molecules of H2 and three molecule of O2 are there in the container Q, 6 molecules of H2 will react with 3 molecules of O2 to produce 3 molecules of H2O. One molecule of Hydrogen will not take part in reaction and will be present in container Q after then reaction, and the mass in overall reaction is conserved!
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Answer:
Pyridine will neutralize the acid by-product of the reaction.
Explanation:
In the esterifications, hydrochloric acid is released as a by-product that reacts directly with the pyridine in the medium to give the pyridine hydrochloride.