Answer:

Explanation:
Hello,
For the given chemical reaction:

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

Finally, we compute the percent yield with the obtained 2.10 g:

Best regards.
Answer:
The answer to your question is V2 = 29.6 l
Explanation:
Data
Pressure 1 = P1 = 12 atm
Volume 1 = V1 = 23 l
Temperature 1 = T1 = 200 °K
Pressure 2 = 14 atm
Volume 2 = V2 = =
Temperature 2 = T2 = 300°K
Process
1.- To solve this problem use the Combine gas law.
P1V1/T1 = P2V2/T2
-Solve for V2
V2 = P1V1T2 / T1P2
2.- Substitution
V2 = (12)(23)(300) / (200)(14)
3.- Simplification
V2 = 82800 / 2800
4.- Result
V2 = 29.6 l
Answer:
Density should be 8.5
Explanation:
When you subtract 17.2-10.0 for water displacement you get 7.2 which is the volume and you divide it by 15.7 so your equation will be 15.7/7.2 = 8.5