Answer:Chemistry problems can be solved using a variety of techniques.
Explanation: Many chemistry teachers and most introductory chemistry texts illustrate problem solutions using the factor-label method. ... The use of analogies and schematic diagrams results in higher achievement on problems involving moles, stoichiometry, and molarity. Hope this helped!
In this case a double displacement reaction will take place.
Answer:
![m_{H_2SO_4}=81.7gH_2SO_4](https://tex.z-dn.net/?f=m_%7BH_2SO_4%7D%3D81.7gH_2SO_4)
![m_{H_2}=1.67gH_2](https://tex.z-dn.net/?f=m_%7BH_2%7D%3D1.67gH_2)
Explanation:
Hello,
Based on the given undergoing chemical reaction is is rewritten below:
![2Al (s) + 3H_2SO_4 (aq)\rightarrow Al _2(SO4)_3 (aq) + 3H_2 (g)](https://tex.z-dn.net/?f=2Al%20%28s%29%20%2B%203H_2SO_4%20%28aq%29%5Crightarrow%20%20Al%20_2%28SO4%29_3%20%28aq%29%20%2B%203H_2%20%28g%29)
By stoichiometry we find the minimum mass of H2SO4 (in g) as shown below:
![m_{H_2SO_4}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2SO_4}{2molAl}*\frac{98gH_2SO_4}{1molH_2SO_4} \\m_{H_2SO_4}=81.7gH_2SO_4](https://tex.z-dn.net/?f=m_%7BH_2SO_4%7D%3D15.0gAl%2A%5Cfrac%7B1molAl%7D%7B27gAl%7D%2A%5Cfrac%7B3molH_2SO_4%7D%7B2molAl%7D%2A%5Cfrac%7B98gH_2SO_4%7D%7B1molH_2SO_4%7D%20%5C%5Cm_%7BH_2SO_4%7D%3D81.7gH_2SO_4)
Moreover, mass of H2 gas (in g) would be produced by the complete reaction of the aluminum block turns out:
![m_{H_2}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2}{2molAl}*\frac{2gH_2}{1molH_2} \\m_{H_2}=1.67gH_2](https://tex.z-dn.net/?f=m_%7BH_2%7D%3D15.0gAl%2A%5Cfrac%7B1molAl%7D%7B27gAl%7D%2A%5Cfrac%7B3molH_2%7D%7B2molAl%7D%2A%5Cfrac%7B2gH_2%7D%7B1molH_2%7D%20%5C%5Cm_%7BH_2%7D%3D1.67gH_2)
Best regards.
The balanced chemical reaction:
C3H8 + 5O2 = 3CO2 + 4H2O
We are given the amount of the carbon dioxide to be produced. This will be the starting point of our calculations.
<span>43.62 L CO2 ( 1 mol CO2 / 22.4 L CO2 ) (5 mol O2 / 3 mol CO2 ) (
22.4 L O2 / 1 mol O2) = 72.7 L O2</span>