Answer:
21.344%
Explanation:
For the given chemical reaction, 8 moles of the reactant should produce 4 moles of . However, 195 g of
was produced instead. The molar mass of
is 61.9789 g/mol.
Thus, the moles of produced = 195/61.9789 = 3.1462 moles
The percent error = [(Actual -Experiment)/Actual]*100%
The percent error = [(4.00 - 3.1462)/4.00]*100% = (0.85376/4.00)*100% = 21.344%
Answer:
The weight percent of potassium carbonate is 50,8 wt% and of potassium bicarbonate 49,2 wt%
Explanation:
The reactions of potassium carbonate (K₂CO₃) and potassium bicarbonate (KHCO₃) with HCl produce:
K₂CO₃ + 2HCl → 2KCl + CO₂ + H₂O
KHCO₃ + HCl → KCl + CO₂ + H₂O
That means that you need 2 moles of HCl to titrate potassium carbonate and 1 mol to titrate potassium bicarbonate.
The moles of HCl to titrate the mixture are:
0,03416L× = <em>0,02603 mol of HCl</em>
If X is mass of K₂CO₃ and Y is mass of KHCO₃ in the mixture, the moles of HCl to titrate the mixture are equals to:
0,02603 mol = 2X× + Y×
<em>(1)</em>
As the mass of the mixture is 2,122g:
2,122g = X + Y <em>(2)</em>
Replacing (2) in (1):
0,02603 mol = 0,01447 (2,122-Y) + 9,988x10⁻³Y
0,02603 mol = 0,0307 - 0,01447Y + 9,988x10⁻³Y
-4,6778x10⁻³ = -4,4827x10⁻³Y
1,044g = Y <em>-mass of potassium bicarbonate-</em>
Thus:
X = 1,078g <em>-mass of potassium carbonate-</em>
The weight percent of potassium carbonate is:
×100 =<em> 50,8 wt%</em>
The weight percent of potassium bicarbonate is:
×100 = <em>49,2 wt%</em>
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I hope it helps!