A 6.00 gram sample of a dry mixture of potassium hydroxide, potassium carbonate, and potassium chloride is reacted with 0.100 liter of 2.0?
A 6.00 gram sample of a dry mixture of potassium hydroxide, potassium carbonate, and potassium chloride is reacted with 0.100 liter of 2.0 molar HCL solution
A 249 milliliter sample of dry CO2 gas, measured at 22*C and 740 mmHg, is obtained from the reaction. What is the percentage of potassium carbonate in the mixture?
- Roger the MoleLv 74 weeks agoBest answer
Supposing the given amount of HCl is sufficient to react with all of the KOH and K2CO3:
n = PV/RT = (740 mmHg) x (0.249 L) / ((62.36367 L mmHg/K mol) x (22 + 273) K) = 0.01002 mol CO2
K2CO3 + 2 HCl → 2 KCl + CO2 + H2O
(0.01002 mol CO2) x (1 mol K2CO3 / 1 mol CO2) x (138.2055 g K2CO3/mol) / (6.00 g) = 0.230803 =
- Anonymous4 weeks ago
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