Answer:
The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL
Explanation:
Using the formula
Ca Va = Cb Vb
Cb = 0.32 M
Vb = 50 mL = 50/1000 = 0.050L
Ca = 0.5 M
Va =?
Substituting for Va in the equation, we obtain:
Va = Cb Vb / Ca
Va = 0.32 * 0.05 / 0.5
Va = 0.016 / 0.5
Va = 0.032 L
The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL
Answer:
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
Explanation:
When we balance a chemical equation, what we are trying to do is to achieve the same number of atoms for each element on both sides of the arrow. On the right of the arrow is where we can find the products, while the reactants are found on the left of the arrow.
We usually balance O and H atoms last.
AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 1
Cl --- 3
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
2 AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of As atoms is now balanced.
2 AsCl₃ + 3 H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of S atoms is now equal on both sides.
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
The equation is now balanced.
The idea here is that you need to figure out how many moles of magnesium chloride,
MgCl
2
, you need to have in the target solution, then use this value to determine what volume of the stock solution would contain this many moles.
As you know, molarity is defined as the number of moles of solute, which in your case is magnesium chloride, divided by liters of solution.
c
=
n
V
So, how many moles of magnesium chloride must be present in the target solution?
c
=
n
V
⇒
n
=
c
⋅
V
n
=
0.158 M
⋅
250.0
⋅
10
−
3
L
=
0.0395 moles MgCl
2
Now determine what volume of the target solution would contain this many moles of magnesium chloride
c
=
n
V
⇒
V
=
n
c
V
=
0.0395
moles
3.15
moles
L
=
0.01254 L
Rounded to three sig figs and expressed in mililiters, the volume will be
V
=
12.5 mL
So, to prepare your target solution, use a
12.5-mL
sample of the stock solution and add enough water to make the volume of the total solution equal to
250.0 mL
.
This is equivalent to diluting the
12.5-mL
sample of the stock solution by a dilution factor of
20
.
To calculate the mass of Fe formed in a) we get first the limiting reactant between Fe2O3 and CO. Given the masses, the ratio of Fe2O3 is 1.33 while that of CO is 1.67. Hence the limiting reagent is Fe2O3. The mass of Fe formed is 148.98 grams. In b, the needed CO is only 112.04 grams. Hence, the excess is 27. 96 grams.
