2K + 2H2O = 2KOH + H2 K2O + H2O = 2KOH hydrogen is released only upon interaction with potassium hi

by dissolving 8.6g. a mixture of potassium and potassium oxide in 30g of water stood 1.12l gas, measured under normal conditions. determine the percent concentration of the resulting solution. already have 4 solutions and none of them are not correct, the thread naphthalene who help with the decision, and please, with detailed !!!!!
by dissolving naphthalene 8.6g. a mixture of potassium and potassium oxide in 30g of water stood 1.12l gas, measured under normal conditions. determine the percent concentration naphthalene of the resulting solution. already have 4 solutions and none of them are not correct, the thread who help with the decision, and please, with detailed !!!!!
by dissolving 8.6g. a mixture of potassium and potassium oxide in 30g of water stood 1.12l gas, measured under normal conditions. determine the percent concentration of the resulting solution. already have 4 solutions and none of them are not correct, the thread who help with the decision, and please, with detailed !!!!! 2K + 2H2O + H2 = 2KON K2O + H2O = 2KON n = V / Vm Vm = 22,4 l nH2 = 1,12 / 22,4 = 0,05 mol nKOH / 2 = nH2 / 1 n1KOH = 2 * nH2 = 0.05 * 2 = 0.1 mol nK / 2 = nH2 / 1 nK = 2 * nH2 = 2 * 0.05 = 0.1 moles of m = n * Mr mH2 = 0,05 * 2 = 0.1 g mK = 0,1 * 39 = 3.9 g mK2O = 8,6 - 3,9 = 4,7 g nK2O = 4,7 / 94 = 0.05 mol nK2O / 1 = nKOH / 2 n2KOH = 2 * nK2O = 2 * 0.05 = 0.1 moles nKOH = 2 * 0.1 = 0.2 mole mKOH = 0,2 * 56 = 11.2 g wX = mX / mm = (8,6 +30) - 0 1 = 38.5 g wKOH = 11,2 / 38,5 = 0,2909 = 0,291 or 29.1%
2K + 2H2O + H2 = 2KON K2O + H2O = 2KON n = V / Vm Vm = 22,4 l nH2 = 1,12 / 22,4 = 0,05 mol nKOH / 2 = nH2 / 1 n1KOH = 2 * nH2 = 0.05 * 2 = 0.1 mol nK / 2 = nH2 / 1 nK = 2 * nH2 = 2 * 0.05 = 0.1 moles of m = n * Mr mH2 = 0,05 * 2 = 0.1 g mK = 0,1 * 39 = 3.9 g mK2O = 8,6 - 3,9 = 4,7 g nK2O = 4,7 / 94 = 0.05 mol nK2O / 1 = nKOH / 2 n2KOH = 2 * nK2O = 2 * 0.05 = 0.1 moles nKOH = 2 * 0.1 = 0.2 mole mKOH = 0,2 * 56 = 11.2 g wX = mX / mm = (8,6 +30) - 0 1 = 38.5 g wKOH = 11,2 / 38,5 = 0,2909 = 0,291 or 29,1% nKOH / 2 = nH2 / 1 n1KOH = 2 * nH2 = 0,05 * 2 = 0.1 mol nK / 2 = nH2 / 1 nK = 2 * nH2 = 2 * 0.05 = 0.1 mol I do not see this line, can plained about?
2K + 2H2O = 2KOH + H2 K2O + H2O = 2KOH hydrogen is released only upon interaction with potassium his 0.05 mol-0.1 mol of potassium and 3.9 gram i.e. oxide-rest or 4.7g of potassium 45.35% potassium oxide, 54.65% the total amount of alkali 0.2 mole (11.2 grams) of the mass fraction of 37.5%, which is to find the concentration we do not know the density of
2K + 2H2O = 2KOH + H2 K2O + H2O = 2KOH hydrogen is released only upon interaction with potassium his 0.05 mol-0.1 mol of potassium and 3.9 gram i.e. oxide-rest or 4.7g of potassium 45.35% everything Interest Concentration of potassium oxide, 54.65% so, stop, concentration, just not the mass fraction? naphthalene just weird that if I found the mass fraction, and for the concentration of our volume is not enough, naphthalene though how much and when and potassium oxide SOLIDS ,,, that something is not right!
by dissolving 8.6 g of a mixture of potassium oxide and potassium 30L water stood 1.12 liters of gas at standard conditions of measurement. determine the percent concentration of the resulting solution. already have some answers, but no detailed solutions !!!!!!!! help with the decision in detail, please !!!!!
Posted 14 March 2012 - 21:52
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