Answer:
H+ + OH− --> H2O
Explanation:
Hydrochloric acid is represented by the chemical formular; HCl. This is an ionic substance so in water it breaks apart into hydrohrn ions; H+ and chloride ions; Cl−. It is a strong acid, hence it completely dissociates.
Potassium Hydroxide is also an ionic substance it also breaks apart in water into potassium ions; K+ and hydroxide ions; OH−. It is a strong base, hence it completely dissociates.
The complete ionic equation for the reaction is given as;
H+ + Cl− + K+ + OH− --> K+ + Cl− + H2O
The Hydrogen ion and the Hydroxide ions combine to form water.
The net ionic equation is given as;
H+ + OH− --> H2O
Cl- and K+ ions were cancelled out because they do not undergo any changes therefore are not part of the net ionic equation. They are referred to as spectator ions.
Answer:
5.83 mol.
Explanation:
- From the balanced reaction:
<em>2Al + 3Ag₂S → 6Ag + Al₂S₃,</em>
It is clear that 2 mol of Al react with 3 mol of Ag₂S to produce 1 mol of Ag and 1 mol of Al₂S₃.
Al reacts with Ag₂S with (2: 3) molar ratio.
<em>So, 2.27 mol of Al reacts completely with 3.4 mol of Ag₂S with (2: 3) molar ratio.</em>
<em />
- The limiting reactant is Ag₂S.
- The excess "left over" reactant is Al.
The reamining moles of excess reactant "Al" = 8.1 mol - 2.27 mol = 5.83 mol.
Answer:
You need to add 400mL of water
Explanation:
500mL = 5 M HCI That means that if you divide both sides by 5
100mL = 1 M HCI If you need ot get rid of 4 M HCI then you add 400 mL of water because that is what it is equal to
Answer:
Explanation:
Approx.
425
⋅
g
Explanation:
2
A
l
(
s
)
+
3
C
l
2
(
g
)
→
2
A
l
C
l
3
(
s
)
You have given a stoichiometrically balanced equation, so bravo.
The equation explicitly tells us that
54
⋅
g
of aluminum metal reacts with
6
×
35.45
⋅
g
C
l
2
gas to give
266.7
⋅
g
of
aluminum trichloride
hope this helps
Answer : The volume of oxygen at STP is 112.0665 L
Solution : Given,
The number of moles of 
= 5 moles
At STP, the temperature is 273 K and pressure is 1 atm.
Using ideal gas law equation :
where,
P = pressure of gas
V = volume of gas
n = the number of moles
T = temperature of gas
R = gas constant = 0.0821 L atm/mole K (Given)
By rearranging the above ideal gas law equation, we get
Now put all the given values in this expression, we get the value of volume.
Therefore, the volume of oxygen at STP is 112.0665 L
