Question

Определите объем аммиака (н.у)образующегося при взаимодействии 8 г.хлорида аммония с избытком гидроксида кальция,если выход продукта реакции по сравнению с теоритическим составил 75%

Answer 1

1)Напишем уравнение реакции:

8 г                           х л

NH4Cl+Ca(OH)2===>2NH3+CaCl2+2H2O(при t=200гр.по Цельсию)

M=53,5 г/моль        Vm=22,4 л/моль

m=53,5 г                 V=44,8л

ВЫводим х= 8г*44,8л/53,5 г=6,7 л

2)Выход продукции по сравнению с теоретическим возможным выходом означает то,что во время опыта(практики) возникли какие-то погрешности и вместо 100% как описанно в теории на практике получилось только 75 % вещества.

а)Мы можем высчитать 75% по формуле: w=m(прак-ая)*100%/m(теор-ая)

Следовательно m(прак-ая)=w*m(теор-ая)/100%=75%*6,7л/100%=5,025 л

б)Мы можем высчитать по пропорции:

6,7л-100%

х л-75%

Следовательно х=75%*6,7л/100%=5,025 л

Ответ:5,025 л

Answer 2

Calculation of Ammonia Volume

To determine the volume of ammonia (NH3) produced when 8 grams of ammonium chloride (NH4Cl) reacts with an excess of calcium hydroxide (Ca(OH)2), we need to consider the balanced chemical equation for the reaction:

2NH4Cl + Ca(OH)2 -> 2NH3 + CaCl2 + 2H2O

From the equation, we can see that 2 moles of NH4Cl react to produce 2 moles of NH3. To find the number of moles of NH4Cl, we can use its molar mass, which is approximately 53.49 g/mol.

Moles of NH4Cl = mass of NH4Cl / molar mass of NH4Cl Moles of NH4Cl = 8 g / 53.49 g/mol

Using the molar ratio from the balanced equation, we can determine the number of moles of NH3 produced:

Moles of NH3 = Moles of NH4Cl

Finally, to calculate the volume of NH3 at standard temperature and pressure (STP), we can use the ideal gas law, which states that 1 mole of any gas occupies 22.4 liters at STP.

Volume of NH3 = Moles of NH3 * 22.4 L/mol

However, we need to take into account that the reaction has a 75% yield, meaning that only 75% of the theoretical amount of NH3 is actually produced.

Volume of NH3 (at STP) = (Moles of NH3 * 22.4 L/mol) * 0.75

Please note that without the exact molar ratio between NH4Cl and NH3, it is not possible to provide an accurate calculation. The given information does not provide the necessary data to determine the molar ratio.