a. 30 moles of H₂O
b. 2.33 moles of N₂
<h3>Further explanation</h3>
Given
a. 20 moles of NH₃
b. 3.5 moles of O₂
Required
a. moles of H₂O
b. moles of N₂
Solution
Reaction
4NH₃+3O₂⇒2N₂+6H₂O
a. From the equation, mol ratio NH₃ : H₂O = 4 : 6, so mol H₂O :
=6/4 x mol NH₃
= 6/4 x 20 moles
= 30 moles
b. From the equation, mol ratio N₂ : O₂ = 2 : 3, so mol N₂ :
=2/3 x mol O₂
= 2/3 x 3.5 moles
= 2.33 moles
Omg im learning this too . i think it’s the third picture ? bc that’s prophase
Currently in this equation, you have 2 hydrogen atoms and 2 oxygen atoms on the left, and then 2 hydrogen atoms and 1 oxygen atom on the right. To balance, you would need to even out the oxygens, so we can first place a 2 in front of H2O to get:
H2 + O2 -> 2H2O
Now, however, you can see that we have too many hydrogen atoms on the right, so to get the final answer, we add a 2 in front of hydrogen on the left:
2H2 + O2 -> 2H2O
I hope this helps!
When the concentration is expressed in mass percentage, that means there is 3 g of solvent H₂O₂ in 100 grams of the solution. Then, that means the remaining amount of solute is 97 g. We use the value of molarity (moles/liters) to determine the amount of solution in liters, denoted as V. The solution is as follows:
0.02 mol KMnO4/L solution * 158.034 g KMnO4/mol * V = 97 g KMnO4
Solving for V,
V = 30.69 L
This equation is balanced
