This
is a neutralization, which means we are mixing a base with an acid until the
mixture becomes neutral. We add more HI to the NaOH until the number of acid
equivalents is equal to the number of base equivalents. We can calculate the
acid equivalents using normality and volume, and the same with base equivalents.
A =
acid equivalents = normality*volume (in the acid solution)
B =
base equivalents = normality*volumen (in the base solution)
A =
B
NA*VA
= NB*VB
HI
is an acid which releases only 1 acid equivalent per molecule, so its molarity
is equal to its normality.
NaOH
is a base which releases only 1 base equivalent per formula unit, so its
molarity is equal to its normality.
MA*VA
= MB*VB
We’re
trying to find out NaOH molarity, which is equal to the NaOH normality.
MB
= MA*VA/VB
DATA:
MA
= 3.0 M
<span>The
volumen of HI used can be calculated by subtracting the
final volume of the burette to its initial volume (the final volume is smaller,
as we have taken some volume away)</span>
VA
= V0-Vf = 0.7 ml - 0.2 ml = 0.5 ml
VB
= V0-Vf = 47.6 ml - 37.5 ml = 10.1 ml
MB
= 3.0M*0.5M/10.1M = 0.149 M
Answer: Theoretical yield is 313.6 g and the percent yield is, 91.8%
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of require 3 moles of
Thus 2.8 moles of will require= of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 1 mole of give = 2 moles of
Thus 2.8 moles of give = of
Mass of
Theoretical yield of liquid iron = 313.6 g
Experimental yield = 288 g
Now we have to calculate the percent yield
Therefore, the percent yield is, 91.8%
Answer:
Option D. 30 g
Explanation:
The balanced equation for the reaction is given below:
2Na + S —> Na₂S
Next, we shall determine the masses of Na and S that reacted from the balanced equation. This is can be obtained as:
Molar mass of Na = 23 g/mol
Mass of Na from the balanced equation = 2 × 23 = 46 g
Molar mass of S = 32 g/mol
Mass of S from the balanced equation = 1 × 32 = 32 g
SUMMARY:
From the balanced equation above,
46 g of Na reacted with 32 g of S.
Finally, we shall determine the mass sulphur, S needed to react with 43 g of sodium, Na. This can be obtained as follow:
From the balanced equation above,
46 g of Na reacted with 32 g of S.
Therefore, 43 g of Na will react with = (43 × 32)/46 = 30 g of S.
Thus, 30 g of S is needed for the reaction.
Answer:
4Fe + 3O₂ → 2Fe₂O₃
Explanation:
Fe → ²⁺
O → ²⁻
But Iron III is Fe³⁺
So we have Fe³⁺ and O²⁻, the formula for the oxide must be Fe₂O₃ so the equation can be:
4Fe + 3O₂ → 2Fe₂O₃