Question

Смесь гидроксида кальция массой 111г и нитрата аммония массой 200г тщательно перемешали и нагрели. Вычислите объем выделившегося газа. (н.у.)

Answer 1

Ca(OH)2+2NH4NO3=Ca(NO3)2+2NH3

n Ca(OH)2=111/74=1.5 моль

n NH4NO3=200/80=2.5 моль

n NH3=2.5 моль

V NH3=2.5*22.4=56 л

 

 

 

 

 

 

 

 

 

 

 

 

 

Answer 2

Calculation of the Volume of Gas Evolved from the Reaction

To calculate the volume of gas evolved from the reaction between calcium hydroxide and ammonium nitrate, we need to determine the balanced chemical equation for the reaction. Once we have the balanced equation, we can use stoichiometry to calculate the volume of gas produced.

The balanced chemical equation for the reaction between calcium hydroxide (Ca(OH)2) and ammonium nitrate (NH4NO3) is as follows:

Ca(OH)2 + 2NH4NO3 → Ca(NO3)2 + 2NH3 + 2H2O.

From the balanced equation, we can see that 1 mole of calcium hydroxide reacts with 2 moles of ammonium nitrate to produce 2 moles of ammonia gas (NH3).

To calculate the volume of gas produced, we need to convert the masses of calcium hydroxide and ammonium nitrate to moles. The molar mass of calcium hydroxide (Ca(OH)2) is 74.09 g/mol, and the molar mass of ammonium nitrate (NH4NO3) is 80.04 g/mol.

Given: - Mass of calcium hydroxide (Ca(OH)2) = 111 g - Mass of ammonium nitrate (NH4NO3) = 200 g

To convert the masses to moles, we use the formula:

moles = mass / molar mass

Calculating the moles of calcium hydroxide: - Moles of Ca(OH)2 = 111 g / 74.09 g/mol = 1.498 mol

Calculating the moles of ammonium nitrate: - Moles of NH4NO3 = 200 g / 80.04 g/mol = 2.499 mol

Since the reaction ratio between calcium hydroxide and ammonium nitrate is 1:2, we can determine that 1.498 moles of calcium hydroxide will react with 2 * 1.498 = 2.996 moles of ammonium nitrate.

According to the stoichiometry of the balanced equation, 2 moles of ammonia gas (NH3) are produced for every 1 mole of calcium hydroxide that reacts. Therefore, the number of moles of ammonia gas produced is also 2.996 mol.

To calculate the volume of gas at standard temperature and pressure (STP), we can use the ideal gas law:

PV = nRT

Where: - P is the pressure (at STP, it is 1 atm) - V is the volume of gas - n is the number of moles of gas - R is the ideal gas constant (0.0821 L·atm/(mol·K)) - T is the temperature (at STP, it is 273.15 K)

Rearranging the equation to solve for V, we get:

V = (nRT) / P

Substituting the values into the equation: - n = 2.996 mol - R = 0.0821 L·atm/(mol·K) - T = 273.15 K - P = 1 atm

Calculating the volume of gas: - V = (2.996 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm = 66.25 L

Therefore, the volume of gas evolved from the reaction is 66.25 liters at standard temperature and pressure (STP).

Please note that this calculation assumes ideal gas behavior and that the reaction goes to completion without any side reactions or losses.

Let me know if there's anything else I can help you with!