Suppose of nickel(II) chloride is dissolved in of a aqueous solution of potassium carbonate. Calculate the final molarity of chl
1 answer:
You might be interested in
Answer:
Explanation:
We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.
Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.
We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.
- Na: 22.9897693 g/mol
- C: 12.011 g/mol
- O: 15.999 g/mol
Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.
- Na₂ = 22.9897693 * 2= 45.9795386 g/mol
- O₃ = 15.999 * 3= 47.997 g/mol
- Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol
We will convert using dimensional analysis. Set up a ratio using the molar mass.
We are converting 57.3 grams to moles, so we multiply by this value.
Flip the ratio so the units of grams of sodium carbonate cancel.
The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.
There are approximately <u>0.541 moles of sodium carbonate</u> in 57.3 grams.
Taking into account definition of percent yield, the percent yield for the reaction is 57.08%.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
CH₄ + Cl₂ → CH₃Cl + HCl
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- CH₄: 1 mole
- Cl₂: 1 mole
- CH₃Cl: 1 mole
- HCl: 1 mole
The molar mass of the compounds is:
- CH₄: 16 g/mole
- Cl₂: 70.9 g/mole
- CH₃Cl: 50.45 g/mole
- HCl: 36.45 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- CH₄: 1 mole ×16 g/mole= 16 grams
- Cl₂: 1 mole ×70.9 g/mole= 70.9 grams
- CH₃Cl: 1 mole ×50.45 g/mole= 50.45 grams
- HCl: 1 mole ×36.45 g/mole= 36.45 grams
Mass of CH₃Cl formed
The following rule of three can be applied: if by reaction stoichiometry 16 grams of CH₄ form 50.45 grams of CH₃Cl, 25 grams of CH₄ form how much mass of CH₃Cl?
<u><em>mass of CH₃Cl= 78.83 grams</em></u>
Then, 78.83 grams of CH₃Cl can be produced from 25 grams of CH₄.
<h3>Percent yield</h3>The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.
The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:
where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.
Percent yield for the reaction in this case
In this case, you know:
- actual yield= 45 grams
- theorical yield= 78.83 grams
Replacing in the definition of percent yields:
Solving:
<u><em>percent yield= 57.08%</em></u>
Finally, the percent yield for the reaction is 57.08%.
Learn more about
the reaction stoichiometry:
percent yield:
#SPJ1