Iron(III) oxide reacts with carbon monoxide according to the equation: Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3 CO₂(g) A reaction mixture initially contains 22.55 g Fe₂O₃ and 14.78 g CO. Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant remains?

Video transcript

Hi everyone today, we have a question asking us at the beginning of the reaction, there are 34.67g of copper oxide and 9.87 g of carbon monoxide. And we want to calculate the amount in grams of excess reactant that remains after the reaction has come to the completion and our reaction is copper copper oxide solid plus three carbon monoxide gashes forms to copper solid plus three carbon carbon dioxide gashes. So let's start out with our three 34 0.67 grams of copper oxide times one mole of copper oxide, divided by 175. g of copper oxide. And our grams here are canceling out, Leaving us with 0. moles of copper oxide. Next we have 9.17 g of carbon monoxide. And we're going to multiply that by one mole of carbon monoxide over .01g of carbon monoxide and our grams of carbon monoxide are going to cancel out. And that's going to leave us with 0.35- moles of carbon monoxide. And for our next step, we want to take our lowest number of moles, which is our 0.1980 and see how many moles of carbon monoxide we get. So we're going to have 0. moles of copper Times 1. 3 moles of carbon monoxide over two moles of copper Times are molar mass of carbon monoxide. So 28.01 g of carbon monoxide over one mole. So our moles of copper are canceling out, and our moles of carbon monoxide are canceling out, and that gives us 8. 1897 grams of carbon monoxide. And our last step is to take our 9.87 g -R 8. 1897, And that equals 1.55 g of carbon monoxide. So our answer here is B. Thank you for watching. Bye.