Alumina is $Al_2O_3$, which is the product of aluminum oxidation. None of your equations shows the correct formulae. We can rewrite the correct equation as: $2Al(s)+ 3/2O_2(g) rarr Al_2O_3(s)$ This...
1 Answers 1 viewsAluminum is oxidized; oxygen is reduced. What are the elemental oxidation states in $Al_2O_3$?
1 Answers 1 views$4Al(s) + 3O_2(g) rarr 2Al_2O_3(g)$ This is a redox reaction, and represents the oxidation of aluminum metal by dioxygen gas to give alumina.
1 Answers 1 viewsTo check excess and limiting reactants, the number of moles has to be found, and then divided by the coefficient. For the sake of shortening the equations, let us say...
1 Answers 1 viewsYou'd need $"33.6 g"$ of aluminium to react with that much manganese dioxide. So, you've got your balanced chemical equation for this reaction $3MnO_(2(s)) + 4Al_((s)) -> 3Mn_((s))...
1 Answers 1 viewsFor propane we have 3 carbons and 8 hydrogen atoms, so the gram-molecular weight is (312 + 81) = 44g/mol. The mole ratio of oxygen to carbon dioxide is 5:3,...
1 Answers 1 views$CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(l)$ The equation explicitly tells us that one equivalent of methane gas reacts with two equiv of dioxygen to give one equiv of carbon...
1 Answers 1 viewsThe limiting reactant is the reactant that is present in less amount. 4 mol of Al need to react completely of 3 mol of Oxygen (according with the reaction you...
1 Answers 1 viewsThis is a limiting reactant problem. We know that we will need a balanced equation and moles of each reactant. 1. Gather all the information in one place with...
1 Answers 1 viewsMass is a measure of the number of particles. Here you have $O$ atoms, and dioxygen, $O_2$, and ozone molecules, $O_3$ but each have the same mass by specification. There...
1 Answers 1 views