Why are science safety rules important? PLZ DO NOT COPY AND PASTE
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You can solve this problem through dimensional analysis.
First, find the molar mass of NaHCO3.
Na = 22.99 g
H = 1.008 g
C = 12.01 g
O (3) = 16 (3) g
Now, add them all together, you end with with the molar mass of NaHCO3.
22.99 + 1.008 + 12.01 + 16(3) = 84.008 g NaHCO3. This number means that for every mole of NaHCO3, there is 84.008 g NaHCO3. In simpler terms, 1 mole NaHCO3 = 84.008 g NaHCO3.
After finding the molar mass of sodium bicarbonate, now you can use dimensional analysis to solve for the number of moles present in 200. g of sodium bicarbonate.
Cross out the repeating units which are g NaHCO3, and the remaining unit is mole NaHCO3
200. * 1 = 200
200/ 84.008 = 2.38
Notice how there are only 3 sig figs in the answer. This is because the given problem only gave three sig figs.
Your final answer is 2.38 mol NaHCO3.
Answer:
- Addition of Ba(OH)2: favors the formation of a precipitate.
- Undergo a chemical reaction forming soluble species.
- Addition of CuSO4 : favors the formation of a precipitate.
Explanation:
Hello,
In this case, since the dissociation reaction of barium sulfate is:
We must analyze the effect of the common ion:
- By adding barium hydroxide, more barium ions will be added to the equilibrium system so the formation of solid barium sulfate will be favored (reaction shifts leftwards towards reactants).
- By adding sodium nitrate, the following reaction will undergo:
So the precipitate will turn into other soluble species.
- By adding copper (II) sulfate, more sulfate ions will be added to the equilibrium system so the formation of solid barium sulfate will be favored (reaction shifts leftwards towards reactants).
All of this is supported by the Le Chatelier's principle.
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Answer:
a) The theoretical yield is 408.45g of
b) Percent yield =
Explanation:
1. First determine the numer of moles of and
.
Molarity is expressed as:
M=
- For the
M=
Therefore there are 1.75 moles of
- For the
M=}{1Lsolution}[/tex]
Therefore there are 2.0 moles of
2. Write the balanced chemical equation for the synthesis of the barium white pigment, :
3. Determine the limiting reagent.
To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:
- For the :
- For the :
As the is the smalles quantity, this is the limiting reagent.
4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.
5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:
Percent yield =
Percent yield =
The real yield is the quantity of barium white pigment you obtained in the laboratory.