
Какой объем газа(н.у.) выделится при сливании раствора, содержащего 2 моль азотной кислоты с
раствором карбоната натрия, содержащим 53г. этой соли?

Ответы на вопрос




Calculating the Volume of Gas Released
To calculate the volume of gas (at standard conditions) released when a solution containing 2 moles of nitric acid is mixed with a solution of sodium carbonate containing 53g of the salt, we can use the stoichiometry of the reaction between nitric acid and sodium carbonate.
The balanced chemical equation for the reaction between nitric acid (HNO3) and sodium carbonate (Na2CO3) is:
2 HNO3 + Na2CO3 → 2 NaNO3 + H2O + CO2
From the balanced equation, we can see that 2 moles of nitric acid react with 1 mole of sodium carbonate to produce 1 mole of carbon dioxide gas.
Using the molar mass of sodium carbonate (Na2CO3), we can calculate the number of moles of sodium carbonate present in 53g of the salt.
The molar mass of Na2CO3 is: - Na: 22.99 g/mol - C: 12.01 g/mol - O: 16.00 g/mol
So, the molar mass of Na2CO3 = 22.99*2 + 12.01 + 16.00*3 = 105.99 g/mol
Therefore, the number of moles of Na2CO3 present in 53g of the salt can be calculated as: moles = mass / molar mass = 53g / 105.99 g/mol
Now, using the stoichiometry of the reaction, we can calculate the volume of carbon dioxide gas produced at standard conditions (0°C and 1 atm pressure) when 2 moles of nitric acid react with the calculated moles of sodium carbonate.
Let's calculate the volume of gas using the given information and the stoichiometry of the reaction.
Volume of gas (at standard conditions) = moles of CO2 * molar volume at STP
The molar volume of gas at standard temperature and pressure (STP) is approximately 22.4 liters per mole.
Calculation:
- Moles of Na2CO3 = 53g / 105.99 g/mol - Moles of CO2 produced = Moles of Na2CO3 (from the balanced equation) - Volume of gas (at standard conditions) = Moles of CO2 * 22.4 L/molLet's calculate the volume of gas using the given information and the stoichiometry of the reaction.


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