if 14.0 g of aluminium reacts with excess sulfuric acid to produce 75.26 g of aluminium sulfate, what is the percent yield?
1 answer:
Taking into account definition of percent yield, the percent yield for the reaction is 84.88%.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
2 Al + 3 H₂SO₄ → Al₂(SO₄)₃ + 3 H₂
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:
- Al: 2 moles
- H₂SO₄: 3 moles
- Al₂(SO₄)₃. 1 mole
- H₂: 3 moles
The molar mass of the compounds is:
- Al: 27 g/mole
- H₂SO₄: 98 g/mole
- Al₂(SO₄)₃: 342 g/mole
- H₂: 2 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Al: 2 moles ×27 g/mole= 54 grams
- H₂SO₄: 3 moles ×98 g/mole= 294 grams
- Al₂(SO₄)₃: 1 mole ×342 g/mole= 342 grams
- H₂: 3 moles ×2 g/mole= 6 grams
The following rule of three can be applied: if by reaction stoichiometry 54 grams of aluminium form 342 grams of aluminium sulfate, 14 grams of aluminium form how much mass of aluminium sulfate?
<u><em>mass of aluminium sulfate= 88.67 grams</em></u>
Then, 88.67 grams of aluminium sulfate can be produced if 14.0 g of aluminium reacts with excess sulfuric acid.
<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.
<h3>Percent yield for the reaction in this case</h3>In this case, you know:
- actual yield= 75.26 grams
- theorical yield= 88.67 grams
Replacing in the definition of percent yields:
Solving:
<u><em>percent yield= 84.88%</em></u>
Finally, the percent yield for the reaction is 84.88%.
Learn more about
the reaction stoichiometry:
percent yield:
#SPJ1
You might be interested in
Answer:
1. 25 moles water.
2. 41.2 grams of sodium hydroxide.
3. 0.25 grams of sugar.
4. 340.6 grams of ammonia.
5. 4.5x10²³ molecules of sulfur dioxide.
Explanation:
Hello!
In this case, since the mole-mass-particles relationships are studied by considering the Avogadro's number for the formula units and the molar mass for the mass of one mole of substance, we proceed as shown below:
1. Here, we use the Avogadro's number to obtain the moles in the given molecules of water:
2. Here, since the molar mass of NaOH is 40.00 g/mol, we obtain:
3. Here, since the molar mass of C6H12O6 is 180.15 g/mol:
4. Here, since the molar mass of ammonia is 17.03 g/mol:
5. Here, since the molar mass of SO2 is 64.06 g/mol:
Best regards!
Answer:- atoms.
Solution:- We have been given the grams of carbon tetrachloride and asked to calculate the number of atoms of chlorine. It is a three step conversion problem. In the first we convert the grams of carbon tetrachloride to moles of it. In second step we convert moles of carbon tetrachloride to moles of chlorine and in the third step we convert the moles of chlorine to atoms of chlorine.
For grams to mole conversion we need the molar mass of the compound. Molar mass of carbon tetrachloride is 153.82 grams per mol. If we look at the formula of carbon tetrachloride then four chlorine are present in it. It means 1 mol of carbon tetrachloride has four moles of chlorine. The calculations are as follows:
= atoms
So, there are atoms in 12.2 grams of
.