
вычеслить сколько объемов хлора можно получить при взаимодеиствии 43,5 г оксида марганца 6 с
концентрированнои соляной кислотой

Ответы на вопрос

43.5 х
MnO2 + 4HCl = MnCl2 + Cl2 + 2H2O
86 22.4
х = 11.3 литров



Calculating the Volume of Chlorine Produced
To calculate the volume of chlorine produced when 43.5 g of manganese oxide (VI) reacts with concentrated hydrochloric acid, we can use the balanced chemical equation for the reaction and the ideal gas law to determine the volume of chlorine gas produced.
The balanced chemical equation for the reaction is: 2MnO2 + 4HCl → 2MnCl2 + 2H2O + Cl2
First, we need to determine the moles of manganese oxide (VI) and then use stoichiometry to find the moles of chlorine produced. Finally, we can use the ideal gas law to calculate the volume of chlorine gas produced.
Step 1: Calculate the Moles of Manganese Oxide (VI)
The molar mass of manganese oxide (VI) (MnO2) is 86.94 g/mol.
Using the given mass of manganese oxide (VI): 43.5 g / 86.94 g/mol = 0.500 moles of MnO2
Step 2: Use Stoichiometry to Find Moles of Chlorine Produced
From the balanced chemical equation, we can see that 2 moles of manganese oxide (VI) produce 1 mole of chlorine gas.
So, the moles of chlorine gas produced: 0.500 moles of MnO2 × (1 mole of Cl2 / 2 moles of MnO2) = 0.250 moles of Cl2
Step 3: Calculate the Volume of Chlorine Gas Produced
The ideal gas law is given by the equation: PV = nRT
Where: P = pressure V = volume n = number of moles R = ideal gas constant T = temperature
Assuming standard temperature and pressure (STP), where T = 273 K and P = 1 atm, and using the ideal gas constant R = 0.0821 L·atm/(K·mol), we can calculate the volume of chlorine gas produced.
V = (0.250 moles) × (0.0821 L·atm/(K·mol)) × (273 K) / (1 atm) = 5.68 L of Cl2
Therefore, when 43.5 g of manganese oxide (VI) reacts with concentrated hydrochloric acid, approximately 5.68 liters of chlorine gas can be produced.


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