This should help :)
Example 1: A 36.0 g sample of water is initially at 10.0 °C.
How much energy is required to turn it into steam at 200.0 °C? (This
example starts with a temperature change, then a phase change followed
by another temperature change.)
Solution:
<span>q = (36.0 g) (90.0 °C) (4.184 J g¯1 °C¯1) = 13,556 J = 13.556 kJ
q = (40.7 kJ/mol) (36.0 g / 18.0 g/mol) = 81.4 kJ
q = (36.0 g) (100.0 °C) (2.02 J g¯1 °C¯1) = 7272 J = 7.272 kJ
q = 102 kJ (rounded to the appropriate number of significant figures)
</span>
Thermal energy isconverted to gas energy
A: Some ice will turn into liquid water.
Answer:
![Y=58.15\%](https://tex.z-dn.net/?f=Y%3D58.15%5C%25)
Explanation:
Hello,
For the given chemical reaction:
![Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)](https://tex.z-dn.net/?f=Al_2S_3%28s%29%20%2B%206%20H_2O%28l%29%20%5Crightarrow%202%20Al%28OH%29_3%28s%29%20%2B%203%20H_2S%28g%29)
We first must identify the limiting reactant by computing the reacting moles of Al2S3:
![n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3](https://tex.z-dn.net/?f=n_%7BAl_2S_3%7D%3D5.00gAl_2S_3%2A%5Cfrac%7B1molAl_2S_3%7D%7B150.158%20gAl_2S_3%7D%20%3D0.0333molAl_2S_3)
Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:
![n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3](https://tex.z-dn.net/?f=n_%7BAl_2S_3%7D%5E%7Bconsumed%7D%3D2.50gH_2O%2A%5Cfrac%7B1molH_2O%7D%7B18gH_2O%7D%2A%5Cfrac%7B1molAl_2S_3%7D%7B6molH_2O%7D%3D0.0231mol%20%20Al_2S_3)
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:
![m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3](https://tex.z-dn.net/?f=m_%7BAl%28OH%29_3%7D%3D0.0231molAl_2S_3%2A%5Cfrac%7B2molAl%28OH%29_3%7D%7B1molAl_2S_3%7D%2A%5Cfrac%7B78gAl%28OH%29_3%7D%7B1molAl%28OH%29_3%7D%20%3D3.61gAl%28OH%29_3)
Finally, we compute the percent yield with the obtained 2.10 g:
![Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%](https://tex.z-dn.net/?f=Y%3D%5Cfrac%7B2.10g%7D%7B3.61g%7D%20%2A100%5C%25%5C%5C%5C%5CY%3D58.15%5C%25)
Best regards.