Question

При взаимодействии 20л (н.у.) газа HCl с газовой смесью,состояшей из азота и аммиака ,объем смеси уменьшился на 25% . Сколько грамм лития полностью вступит в реакцию с азотом ,имеющимся в смеси ,если на сгорание оставшейся части газовой смеси затрачено 15л(н.у.) кислорода? Ar(Li)=7

Answer 1

20л    20л
HCI+ NH3→NH4CI
1л      1л
С хлороводородом взаимодействует только аммиак,найдем общий объем смеси
25%---------20л
100%------хл   х=80 литров
С кислородом тоже реагирует только аммиак
    ул    15л
4NH3+3О2→2N2+6Н2О
4л         3л                           у=20 литров
значить в сходной смеси было 40л (20+20=40л) амиака и
 80-40=40 литров азота
40л        аг
N2   +   6Li→2Li3N  a=75 г
22,4л     42г
ответ 75 г лития

Answer 2

Calculation of the Reaction:

To calculate the amount of lithium that will react with the nitrogen present in the gas mixture, we need to consider the stoichiometry of the reaction between lithium and nitrogen. The balanced equation for the reaction is:

2Li + N2 -> 2Li3N

From the equation, we can see that 2 moles of lithium react with 1 mole of nitrogen to produce 2 moles of lithium nitride.

Given that the volume of the gas mixture decreased by 25% after the reaction, we can assume that the number of moles of gas remains constant. Therefore, the initial number of moles of nitrogen in the gas mixture is the same as the final number of moles of lithium nitride.

To find the number of moles of nitrogen, we can use the ideal gas law equation:

PV = nRT

Where: P = pressure (assume constant) V = volume of the gas mixture n = number of moles of nitrogen R = ideal gas constant T = temperature (assume constant)

Since the pressure, temperature, and volume are constant, we can simplify the equation to:

n = PV / RT

Now, we can substitute the given values into the equation to find the number of moles of nitrogen.

Next, we need to convert the number of moles of nitrogen to grams using the molar mass of nitrogen (N2).

Finally, we can calculate the amount of lithium that will react with the nitrogen by using the stoichiometry of the reaction.

Let's perform the calculations step by step.

Calculation Steps:

Step 1: Calculate the number of moles of nitrogen in the gas mixture. - Given volume of the gas mixture: 20 L (n.u.) - Given volume of oxygen used for combustion: 15 L (n.u.)

Step 2: Calculate the number of moles of nitrogen using the ideal gas law. - Assume constant pressure, temperature, and volume. - Use the ideal gas constant R = 0.0821 L·atm/(mol·K). - Convert the given volumes to liters (n.u. = normal units).

Step 3: Convert the number of moles of nitrogen to grams using the molar mass of nitrogen (N2). - The molar mass of nitrogen (N2) is approximately 28 g/mol.

Step 4: Calculate the amount of lithium that will react with the nitrogen using the stoichiometry of the reaction. - From the balanced equation: 2 moles of Li react with 1 mole of N2.

Let's perform the calculations now.

Calculation:

Step 1: Calculate the number of moles of nitrogen in the gas mixture. - Given volume of the gas mixture: 20 L (n.u.) - Given volume of oxygen used for combustion: 15 L (n.u.)

The volume of the gas mixture decreased by 25% after the reaction. Therefore, the final volume of the gas mixture is 75% of the initial volume.

Final volume of the gas mixture = 0.75 * 20 L = 15 L

Step 2: Calculate the number of moles of nitrogen using the ideal gas law. - Assume constant pressure, temperature, and volume. - Use the ideal gas constant R = 0.0821 L·atm/(mol·K). - Convert the given volumes to liters (n.u. = normal units).

n = PV / RT

n = (15 L) * (1 atm) / (0.0821 L·atm/(mol·K) * T)

Since the temperature (T) is not given, we cannot calculate the exact number of moles of nitrogen without this information.

Step 3: Convert the number of moles of nitrogen to grams using the molar mass of nitrogen (N2). - The molar mass of nitrogen (N2) is approximately 28 g/mol.

Without the exact number of moles of nitrogen, we cannot calculate the exact mass of nitrogen without the temperature information.

Step 4: Calculate the amount of lithium that will react with the nitrogen using the stoichiometry of the reaction. - From the balanced equation: 2 moles of Li react with 1 mole of N2.

Without the exact mass of nitrogen, we cannot calculate the exact amount of lithium that will react with the nitrogen.

Please provide the temperature (T) or any additional information required to complete the calculation.