Strong acid-titrated with strong base. Suppose the titration was reversed in question 2. If your titrated 30.0 mL of 0.1 M HCl w
1 answer:
You might be interested in
Answer: .
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of K = 49.4 g
Mass of S = 20.3 g
Mass of O = 30.3 g
Step 1 : convert given masses into moles.
Moles of K=
Moles of S= \frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{20.3g}{32g/mole}=0.63moles[/tex]
Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{30.3g}{16g/mole}=1.89moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For K =
For S =
For O =
The ratio of K: S:O = 2: 1: 3
Hence the empirical formula is .
The empirical formula of the substance would be
The constituents of the compound are as follows:
N = 54/201.6 = 26.78% = 26.78/14 = 1.91 moles
H = 15/201.6 = 7.44% = 7.44/1 = 7.44 moles
Cl = 132.6/201.6 = 65.48% = 65.48/35.5 = 1.84 moles
Divide the number of moles by the smallest"
N = 1.91/1.84 = 1
H = 7.44/1.84 = 4
Cl = 1.84/1.84 = 1
Thus, the empirical formula is
More on empirical formula can be found here: brainly.com/question/14044066
#SPJ1