Question

How many grams of o2 are consumed to precipitate all of the iron in 55.0 ml of 0.0350 m fe(ii)?

Answer 1

To find the grams of O2 consumed to precipitate all of the iron in the given solution, use stoichiometry to convert moles of Fe(II) to moles of O2 and then convert moles of O2 to grams.

Step-by-step explanation:

To determine the grams of O2 consumed to precipitate all of the iron in 55.0 mL of 0.0350 M Fe(II), we need to use stoichiometry. First, we calculate the moles of Fe(II) using the molarity and volume of the solution. Then, we use the balanced chemical equation to find the mole ratio between Fe(II) and O2. Finally, we convert the moles of O2 to grams using the molar mass of O2.

Answer 2

Rust forms when 2 atoms of fe reacts with oxygen to form iron oxide as follows. 4Fe2 + 302 ==[]::::::::::::::::> 2Fe2O3. So basically 4 atoms of oxygen reacts with 3 atoms of oxygen to produce the product. We need to find the number of moles of Fe that reacts with Oxygen. Number of moles is the concentration of iron multiplied by the molarity. So we have 55 * 0.035 = 1.925 moles. Hence 1.925*4 atoms of iron which gives 7.7 moles will actually react with 3 *1.925 = 5.775 of Oxygen Since we know the number of moles of oxygen we can solve for its mass. No of moles = mass/molar mass. Mass = no of moles * molar mass = 5.775 * 32 = 184.5