Question

Смесь равных обьемов азота и водорода общим обьемом 11,2 дома превращается в аммиак. Для нейтрализации полученого аммиака нужно 24 см. соляной кислоты (w=9,125%, p=1,04г/см³). Найдите объёмы газов в газовой смеси на выходе после катализатора.

Answer 1

Ответ:

Аммиак 1,4 л

Азот 4,9 л

Водород 3,5 л

Объяснение:

Масса раствора соляной кислоты

24*1,04=24,96 г

Масса кислоты в нём

24,96*0,09125=2,2776 г

Количество хлороводорода

2,2776/36,5=0,0624 моль

HCl+NH3=NH4Cl

Объём полученного амммиака

0,0624*22,4=1,4 л

N2+3H3=2NH3

0,7    2,1     1,4

Объём вступившего в реакцию азота

1,4/2*1=0,7 л

Объём непрореагировшего азота

11,2/2-0,7=4,9 л

Объём водорода, вступившего в реакцию

1,4*3/2=2,1 л

Объём водорода, оставшегося в смеси после реакции

11,2/2-2,1=3,5 л

Answer 2

#### Calculation of Gas Volumes To find the volumes of gases in the gas mixture after the catalyst, we need to use the given information about the initial gas mixture and the amount of hydrochloric acid required for neutralization. Given: - The initial gas mixture consists of equal volumes of nitrogen and hydrogen with a total volume of 11.2 L. - 24 cm³ of hydrochloric acid (HCl) is required for neutralization. - The concentration of hydrochloric acid is 9.125% (w/v), which means it contains 9.125 g of HCl per 100 mL of solution. - The density of hydrochloric acid is 1.04 g/cm³. To solve this problem, we can follow these steps: 1. Calculate the amount of hydrochloric acid in grams using the given concentration and volume. 2. Convert the amount of hydrochloric acid to moles using its molar mass. 3. Use the balanced chemical equation for the reaction between ammonia (NH₃) and hydrochloric acid (HCl) to determine the stoichiometric ratio. 4. Calculate the moles of ammonia produced. 5. Use the ideal gas law to calculate the volume of ammonia gas produced. Let's perform these calculations step by step. #### Step 1: Calculate the amount of hydrochloric acid in grams The concentration of hydrochloric acid is given as 9.125% (w/v), which means it contains 9.125 g of HCl per 100 mL of solution. Since we have 24 cm³ of hydrochloric acid, we need to convert it to grams. **Calculation:** 24 cm³ * (1 mL / 1 cm³) * (9.125 g / 100 mL) = 2.19 g Therefore, the amount of hydrochloric acid required for neutralization is 2.19 grams. #### Step 2: Convert the amount of hydrochloric acid to moles To convert grams to moles, we need to know the molar mass of hydrochloric acid (HCl). The molar mass of HCl is approximately 36.46 g/mol. **Calculation:** 2.19 g / 36.46 g/mol = 0.060 moles Therefore, the amount of hydrochloric acid in moles is 0.060 moles. #### Step 3: Determine the stoichiometric ratio The balanced chemical equation for the reaction between ammonia (NH₃) and hydrochloric acid (HCl) is as follows: NH₃ + HCl → NH₄Cl From the balanced equation, we can see that 1 mole of ammonia reacts with 1 mole of hydrochloric acid to produce 1 mole of ammonium chloride. #### Step 4: Calculate the moles of ammonia produced Since the stoichiometric ratio is 1:1, the moles of ammonia produced will also be 0.060 moles. #### Step 5: Calculate the volume of ammonia gas produced To calculate the volume of ammonia gas produced, we can use the ideal gas law, which states: PV = nRT Where: - P is the pressure of the gas (assumed to be constant) - V is the volume of the gas - n is the number of moles of the gas - R is the ideal gas constant (0.0821 L·atm/(mol·K)) - T is the temperature in Kelvin Since the pressure and temperature are not given, we can assume standard conditions (1 atm and 273 K). **Calculation:** V = (nRT) / P = (0.060 mol * 0.0821 L·atm/(mol·K) * 273 K) / 1 atm = 1.42 L Therefore, the volume of ammonia gas produced after the catalyst is 1.42 liters. Please note that the calculations provided above are based on the given information and assumptions.