Question

При прокаливании оксида марганца IV (MnO2) образуется оксид состава Mn3O4 и выделяется кислород. Вычислите массу оксида марганца IV, который нужно прокалить, чтобы получить 0,5 моль кислорода

Answer 1

Решение на прикрепленной фотографии

Answer 2

Calculation of the Mass of Manganese Dioxide (MnO2) Needed to Produce 0.5 Moles of Oxygen

To calculate the mass of manganese dioxide (MnO2) needed to produce 0.5 moles of oxygen, we need to use the balanced chemical equation for the reaction. The balanced equation for the reaction is as follows:

2 MnO2 -> Mn3O4 + O2

From the balanced equation, we can see that 2 moles of MnO2 produce 1 mole of O2. Therefore, 1 mole of MnO2 will produce 0.5 moles of O2.

To calculate the mass of MnO2 needed, we can use the molar mass of MnO2, which is 86.94 g/mol.

Step 1: Calculate the molar mass of MnO2: Molar mass of Mn = 54.94 g/mol Molar mass of O = 16.00 g/mol Molar mass of MnO2 = (54.94 g/mol) + 2(16.00 g/mol) = 86.94 g/mol

Step 2: Calculate the mass of MnO2 needed: Mass of MnO2 = (0.5 moles of O2) * (86.94 g/mol) = 43.47 g

Therefore, the mass of manganese dioxide (MnO2) needed to produce 0.5 moles of oxygen is approximately 43.47 grams.

Please note that the sources provided did not directly provide the answer to the question. However, the calculation is based on the balanced chemical equation and the molar mass of MnO2, which are fundamental concepts in chemistry.