NH4Cl + NaOH = ? уравнение реакции Рассчитайте массу хлорида натрия, полученного растворением 10 г хлорида аммония в гидроксиде натрия (350 мл 30%-ного раствора плотностью 1,3279 г/мл) (NH4Cl + NaOH = ?). Определите, какое из исходных веществ было в избытке? Запишите молекулярное и ионное уравнение реакции.

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Molecular equation: NH4Cl + NaOH = NaCl + NH3 + H2O
Ionic equation: NH4+ + Cl- + Na+ + OH- = Na+ + Cl- + NH3 + H2O

First, let's calculate the molar mass of NaCl:
Na = 22.99 g/mol
Cl = 35.45 g/mol
NaCl = 22.99 + 35.45 = 58.44 g/mol

Now, let's calculate the amount of NaOH in the solution:
Amount of NaOH = 350 ml x 0.30 x 1.3279 g/ml = 139.19 g
Moles of NaOH = 139.19 g / 40.00 g/mol = 3.48 mol

Next, let's calculate the amount of NH4Cl used in the reaction:
Moles of NH4Cl = 10 g / 53.49 g/mol = 0.19 mol

Since NaOH is in excess, we need to find the limiting reactant, which is NH4Cl in this case.

Now, let's calculate the mass of NaCl produced:
Moles of NaCl = 0.19 mol
Mass of NaCl = 0.19 mol x 58.44 g/mol = 11.104 g

Therefore, 11.104 grams of sodium chloride are produced by dissolving 10 grams of ammonium chloride in a 350 ml solution of 30% sodium hydroxide.

NH4Cl was in excess in this reaction.