Molarity = mol/liter
0.708M = 0.098mol/L
Rearrange to find L:
0.098mol/0.708M = .138L
For every liter there is 1000 mL:
.138L • 1000mL =138mL KOH
Answer:
It is already balanced equation
Answer:
90.3 L
Explanation:
Given data:
Volume of water produced = 77.4 L
Volume of oxygen required = ?
Solution:
Chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
It is known that,
1 mole = 22.414 L
There are 7 moles of oxygen = 7×22.414 = 156.9 L
There are 6 moles of water = 6×22.414 = 134.5 L
Now we will compare:
H₂O : O₂
134.5 : 156.9
77.4 : 156.9/134.5×77.4 =90.3 L
So for the production of 77.4 L water 90.3 L oxygen is required.
2KMnO4+3Na2SO3+H2O→2MnO2+3Na2SO4+2KOH
In a reaction, the reducing agent is the element or compound that donates electron or the one tht loses electrons. The oxidized species. The opposite is called the oxidizing agent. It is the one who accepts the electrons lost. For this reaction KMnO4 is reduced into MnO2.
