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

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

4hours and 17 mins i think

Adozen is 1221/12 = 1.751.75 x 12 = 21therefore, 1.75 is the answer.