Question

найти обьем газа, который образовался при взаимодействии концентрированной азотной кислотой и медею(2) , если масса кислоты = 2,4 кг, а масса металла = 1,9​

Answer 1

Ответ:

Объяснение:

найти объем газа, который образовался при взаимодействии концентрированной азотной кислотой и медью(2) , если масса кислоты = 2,4 кг, а масса металла = 1,9​

    1,9 кг      2,4 кг

      Сu   +  4HNO3   =  Cu(NO3)2  +  2NO2  +  2H2O  

 Молярная  масса  азотной  кислоты  63 г/ моль

 Находим  количество  вещества  азотной  кислоты:

n(HNO3) =   2,4  кг :  63 г/моль =  0,038 кмоль.   Следовательно, (смотрите уравнение)  меди  будет  затрачено  в 4  раза  меньше  

 n(Cu) =  0,038 кмоль : 4  =   0,0095 кмоль.  Медь  взята   избытке.  Всего  меди  вступит  в реакцию : 64 г/моль × 0,0095 кмоль = 0,608  кг  

Если  количество  азотной  кислоты  0,038  кмоль,  то газа  диоксида  азота  образуется  в 2  раза  больше:   0,038  ×2 = 0,076 кмоль

 Вычислим  объем  газа:  22,4 л/моль  ×  0,076 кмоль =  1,7 м³

Answer 2

#### Calculation of Gas Volume Produced To find the volume of gas produced when concentrated nitric acid reacts with copper, we need to use the balanced chemical equation for the reaction. The balanced equation for the reaction between nitric acid (HNO3) and copper (Cu) is as follows: 2 HNO3 + 3 Cu → 3 Cu(NO3)2 + 2 NO + H2O From the balanced equation, we can see that 2 moles of nitric acid react with 3 moles of copper to produce 2 moles of nitric oxide gas (NO). Given: - Mass of nitric acid (HNO3) = 2.4 kg - Mass of copper (Cu) = 1.9 kg To find the volume of gas produced, we need to convert the masses of the substances to moles using their molar masses and then use the mole ratio from the balanced equation to calculate the moles of gas produced. Finally, we can use the ideal gas law to calculate the volume of gas. 1. Calculate the moles of nitric acid (HNO3): - Molar mass of HNO3 = 63.01 g/mol - Moles of HNO3 = (mass of HNO3) / (molar mass of HNO3) - Moles of HNO3 = (2.4 kg * 1000 g/kg) / (63.01 g/mol) 2. Calculate the moles of copper (Cu): - Molar mass of Cu = 63.55 g/mol - Moles of Cu = (mass of Cu) / (molar mass of Cu) - Moles of Cu = (1.9 kg * 1000 g/kg) / (63.55 g/mol) 3. Calculate the moles of nitric oxide gas (NO) produced: - According to the balanced equation, the mole ratio between HNO3 and NO is 2:2. - Moles of NO = (moles of HNO3) * (2 moles of NO / 2 moles of HNO3) 4. Calculate the volume of gas using the ideal gas law: - The ideal gas law equation is: PV = nRT - We need to solve for V (volume), so rearrange the equation: V = (nRT) / P - Assuming standard temperature and pressure (STP), where T = 273.15 K and P = 1 atm. - R is the ideal gas constant, which is 0.0821 L·atm/(mol·K). **Please note that the calculation provided below is an example and may not reflect the actual values for this specific reaction.** Let's calculate the volume of gas produced using the given masses of nitric acid and copper: 1. Calculate the moles of nitric acid (HNO3): - Molar mass of HNO3 = 63.01 g/mol - Moles of HNO3 = (2.4 kg * 1000 g/kg) / (63.01 g/mol) = 38.09 mol 2. Calculate the moles of copper (Cu): - Molar mass of Cu = 63.55 g/mol - Moles of Cu = (1.9 kg * 1000 g/kg) / (63.55 g/mol) = 29.89 mol 3. Calculate the moles of nitric oxide gas (NO) produced: - According to the balanced equation, the mole ratio between HNO3 and NO is 2:2. - Moles of NO = (38.09 mol) * (2 mol NO / 2 mol HNO3) = 38.09 mol 4. Calculate the volume of gas using the ideal gas law: - Assuming standard temperature and pressure (STP), where T = 273.15 K and P = 1 atm. - R = 0.0821 L·atm/(mol·K) - V = (nRT) / P = (38.09 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm = 876.38 L Therefore, the volume of gas produced when concentrated nitric acid reacts with copper is approximately **876.38 liters**.