Question

Какой объем аммиака(н.у.) выделится при нагревании 13,2 г сульфата аммония с гидроксидом натрия(избыток)? Выход реакции 75%. ПОЖАЛУЙСТА ПОМОГИТЕ

Answer 1

(NH₄)₂SO₄+2NaOH=Na₂SO₄+2H₂O+2NH₃↑

m((NH₄)₂SO₄)=m/M=13.2/132=0.1моль

Если смотреть по 100%-ому выходу реакции, то получается вдвое больше 0.1•2=0.2моль аммика. 75%-ов 0.2моль будет нашим искомым аммиаком.

(0.2•75)/100=0.15моль

получим ответ в литрах

V(NH₃)=n•Vm=0.15•22.4=3.36л по 75%-ому выходу реакции

Answer 2

Calculation of Ammonia Volume

To calculate the volume of ammonia gas (at standard temperature and pressure) that will be produced when 13.2 grams of ammonium sulfate reacts with an excess of sodium hydroxide, we need to consider the balanced chemical equation for the reaction. The reaction can be represented as follows:

2(NH4)2SO4 + 2NaOH → 2NH3 + Na2SO4 + 2H2O

From the balanced equation, we can see that 2 moles of ammonium sulfate react to produce 2 moles of ammonia gas. Therefore, we need to convert the given mass of ammonium sulfate to moles and then use the stoichiometry of the reaction to determine the moles of ammonia gas produced.

The molar mass of ammonium sulfate ([(NH4)2SO4]) is calculated as follows: 2(N) + 8(H) + 1(S) + 4(O) = 2(14.01 g/mol) + 8(1.01 g/mol) + 1(32.07 g/mol) + 4(16.00 g/mol) = 132.14 g/mol

To convert the mass of ammonium sulfate to moles, we use the formula: moles = mass / molar mass

moles of ammonium sulfate = 13.2 g / 132.14 g/mol = 0.1 mol

Since the reaction has a 75% yield, we need to multiply the moles of ammonia gas by the yield to find the actual moles of ammonia gas produced: moles of ammonia gas = 0.1 mol * 0.75 = 0.075 mol

Finally, we can convert the moles of ammonia gas to volume using the ideal gas law, which states that: volume = moles * molar volume

The molar volume of a gas at standard temperature and pressure (STP) is 22.4 L/mol.

volume of ammonia gas = 0.075 mol * 22.4 L/mol = 1.68 L

Therefore, approximately 1.68 liters of ammonia gas (at standard temperature and pressure) will be produced when 13.2 grams of ammonium sulfate reacts with an excess of sodium hydroxide, assuming a 75% yield.

Please note that the above calculation assumes ideal conditions and may not account for other factors that could affect the actual yield of the reaction.