KOH+ HNO3--> KNO3+ H2O<span>
From this balanced equation, we know that 1 mol
HNO3= 1 mol KOH (keep in mind this because it will be used later).
We also know that 0.100 M KOH aqueous
solution (soln)= 0.100 mol KOH/ 1 L of KOH soln (this one is based on the
definition of molarity).
First, we should find the mole of KOH:
100.0 mL KOH soln* (1 L KOH soln/
1,000 mL KOH soln)* (0.100 mol KOH/ 1L KOH soln)= 1.00*10^(-2) mol KOH.
Now, let's find the volume of HNO3 soln:
1.00*10^(-2) mol KOH* (1 mol HNO3/ 1 mol KOH)* (1 L HNO3 soln/ 0.500 mol HNO3)* (1,000 mL HNO3 soln/ 1 L HNO3 soln)= 20.0 mL HNO3 soln.
The final answer is </span>(2) 20.0 mL.<span>
Also, this problem can also be done by using
dimensional analysis.
Hope this would help~
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Answer:
Percentage by mass of oxygen = 76.20% (Approx)
Explanation:
Given:
HNO3
H=1, N=14, O=16]
Find:
Percentage by mass of oxygen
Computation:
HNO3
Total mass = 1 + 14 + 3(16)
Total mass = 63
Mass of oxygen = (3)(16) = 48
Percentage by mass of oxygen = [48/63]100
Percentage by mass of oxygen = 76.20% (Approx)
