The question is: What is the mass copper metal that can be prepared from the reaction of 1.25 g of aluminum metal with excess CuSO4 according to the following reaction: 2Al(s)+3CuSO4(aq)-----3Cu(s)+Al2(SO4)3(aq)The answer is mol of Al0.04633 mol, mol of Cu0.06949 mol, grams of Cu4.42How was this figured? I am studing for a final and all the instructor gave us was a review sheet with qa. I have no idea how or what these questions are figured. any help would be great.
This is all about stoichiometry. I'm not going to do all the math, but here's how to do it. * Look at the balanced equation and see that 2 moles Al(s) and 3 moles CuSO4 makes 3 moles Cu(s) and 1 mole Al2(SO4)3. So, always work it mole. * Next, to get the mass of Cu metal that is formed from 1.25 g of Al metal, you convert the 1.25 grams of Al to moles of Al. Then knowing that 2 moles Al produces 3 moles Cu, you can then calculate moles of Cu that can be formed. Note that CuSO4 is in excess, meaning that ALL the Al(s) will react. * Now that you have the moles of Cu metal calculated, you can convert that to mass of Cu(s) by using the atomic weight of Cu.
