Question

Визначте обєм газу який виділяється в наслідок прожарювання 250 грам технічного кальцій карбонату з масовою часткою домішок 15%

Answer 1

СаСО3⇒СаО+СО2↑ определим чистый реагент  100-15=85% m(CaCO3)=250×0,85=212,5г перейдём к количеству n(CaCO3)=212,5/(40+12+48)=2,125моль, согласно уравнения газа образуется столько же и тогда его объём V=2,125×22,4=47,6литра.

Answer 2

Дано
m(технCaCO3)=250 g
w(прим)=15%
----------------------
V(CO2)-?
m(чистCaCO3)=250-(250*15%/100%)=212.5 g
 212.5                 XL
CaCO3-->CaO+CO2
M(CaCO3)=100 g/mol
n(CaCO3)=m/M=212.5/100=2.125 mol
n(CaCO3)=n(CO2)=2.125 mol
V(CO2)=n*Vm=2.125*22.4=47.6 L
ответ 47.6 л

Answer 3

Calculation of Gas Volume Released from the Combustion of 250g of Technical Calcium Carbonate with a 15% Impurity Content

To determine the volume of gas released during the combustion of 250 grams of technical calcium carbonate with a 15% impurity content, we need to consider the chemical reaction involved. The combustion of calcium carbonate produces carbon dioxide gas (CO2) as one of the products.

The balanced chemical equation for the combustion of calcium carbonate is as follows:

CaCO3(s) → CaO(s) + CO2(g)

According to the equation, one mole of calcium carbonate (CaCO3) produces one mole of carbon dioxide (CO2). To calculate the volume of gas released, we need to convert the mass of calcium carbonate to moles and then determine the volume of carbon dioxide gas produced.

Step 1: Convert Mass to Moles

To convert the mass of calcium carbonate to moles, we need to know the molar mass of calcium carbonate. The molar mass of calcium carbonate (CaCO3) is calculated by summing the atomic masses of its constituent elements:

- Atomic mass of calcium (Ca) = 40.08 g/mol - Atomic mass of carbon (C) = 12.01 g/mol - Atomic mass of oxygen (O) = 16.00 g/mol

Molar mass of calcium carbonate (CaCO3) = (40.08 g/mol) + (12.01 g/mol) + (3 * 16.00 g/mol) = 100.09 g/mol

Now, we can calculate the number of moles of calcium carbonate:

Number of moles of calcium carbonate = Mass of calcium carbonate / Molar mass of calcium carbonate

Number of moles of calcium carbonate = 250 g / 100.09 g/mol

Step 2: Calculate Volume of Carbon Dioxide Gas

Since the balanced chemical equation tells us that one mole of calcium carbonate produces one mole of carbon dioxide, the number of moles of carbon dioxide gas released will be the same as the number of moles of calcium carbonate.

Number of moles of carbon dioxide gas = Number of moles of calcium carbonate

Now, we can use the ideal gas law to calculate the volume of carbon dioxide gas:

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 (assumed to be constant)

Since the pressure and temperature are assumed to be constant, we can rearrange the equation to solve for the volume:

V = nRT / P

Substituting the values:

Volume of carbon dioxide gas = (Number of moles of carbon dioxide gas) * (R) * (T) / P

Please provide the temperature and pressure conditions under which the combustion is taking place, so we can calculate the volume of gas released accurately.