Question

К раствору нитрата аммония объемом 90,1 см3 (ω = 12,0 %; ρ = 1,11 г/см3 ) добавили раствор KOH массой 75 г (ω = 25,0 %). Раствор выпарили, остаток прокалили. Рассчитайте массу веществ в твердом остатке после прокаливания.

Answer 1

Решение.
m(NH4NO3) = 90.1 см³*1.11 г/см³*0.12 = 12 г
M(NH4NO3) = 80 г/моль
n(NH4NO3) = 12 г/80 г/моль = 0,15 моль
m(KOH) = 75 г*0.25 = 18,75 г
n(KOH) = 18.75 г/56 г/моль = 0,33 моль

NH4NO3 + KOH = KNO3 + NH3↑ + H2O
Из уравнения реакции следует, что 
n(NH4NO3)=n(KOH), а по расчетам исходя из условия задачи
nNH4NO3)<n(KOH), т.е. КОН взят в избытке.
n(KNO3)=n(NH4NO3)=0,15 моль
2KNO3 =t°⇒2KNO2 + O2↑
M(KNO2) = 85 г/моль
m(KNO2) = 0,15 моль*85 г/моль = 12,75 г
Останется неизрасходованный КОН в количестве 0,33-0,15 = 0,18 мольm(KOH) = 0.18моль*56 г/моль = 10,08 гВсего в сухом остатке после прокаливания: 12,75 г+10,08 г = 22,83 гw(KOH) = 10.08/22.83 = 44.15%w(KNO2) = 12.75/22.83 = 55.85%Ответ: 12,75 г KNO2 и 10,08 г КОН

Answer 2

Calculation of the Mass of Substance in the Solid Residue after Ignition

To calculate the mass of the substance in the solid residue after ignition, we need to consider the reaction between ammonium nitrate (NH4NO3) and potassium hydroxide (KOH). The balanced equation for this reaction is as follows:

NH4NO3 + KOH → NH3 + H2O + KNO3

From the given information, we know the volume of the ammonium nitrate solution is 90.1 cm3 with a concentration of 12.0% (ω) and a density of 1.11 g/cm3. We also know that the mass of the potassium hydroxide solution added is 75 g with a concentration of 25.0% (ω).

To calculate the mass of the substance in the solid residue, we need to determine the limiting reactant in the reaction. The limiting reactant is the one that is completely consumed and determines the amount of product formed.

To find the limiting reactant, we can compare the number of moles of ammonium nitrate and potassium hydroxide. The molar mass of ammonium nitrate (NH4NO3) is 80.04 g/mol, and the molar mass of potassium hydroxide (KOH) is 56.11 g/mol.

Let's calculate the number of moles of ammonium nitrate and potassium hydroxide:

Number of moles of ammonium nitrate: n(NH4NO3) = (mass of NH4NO3) / (molar mass of NH4NO3) n(NH4NO3) = (volume of solution) x (density of solution) x (concentration of NH4NO3) / (molar mass of NH4NO3) n(NH4NO3) = (90.1 cm3) x (1.11 g/cm3) x (12.0% / 100) / (80.04 g/mol)

Number of moles of potassium hydroxide: n(KOH) = (mass of KOH) / (molar mass of KOH) n(KOH) = (75 g) x (25.0% / 100) / (56.11 g/mol)

Now, we can compare the moles of ammonium nitrate and potassium hydroxide to determine the limiting reactant.

Next, we need to calculate the number of moles of the substance in the solid residue after ignition. According to the balanced equation, the stoichiometric ratio between ammonium nitrate and the solid residue is 1:1. This means that for every mole of ammonium nitrate reacted, we will have one mole of the substance in the solid residue.

Finally, we can calculate the mass of the substance in the solid residue by multiplying the number of moles of the substance by its molar mass.

Please note that the exact calculation requires the molar mass of the substance in the solid residue, which is not provided in the given information. Without this information, we cannot provide the precise mass of the substance in the solid residue after ignition.

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