Since 8x3=24 and since 12x3=36 you need to add 6oz of chemical B to have 24oz of chemical A and 36oz of chemical B. Take 1/3 of the total mixture...
1 Answers 1 viewsAlumina 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 viewsI think you would immediately recognize that you had been short-changed. Now looking at the equation, $2Al + 3O_2 rarr 2Al_2O_3$, on each side there are 6 oxygen...
1 Answers 1 viewsZerovalent iron is oxidized to give $Fe(II^+)$. Zerovalent oxygen is reduced from $0$ to $-II$. Each year significant amounts of money are spent to prevent the corrosion of structural steel....
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 viewsAluminum metal is oxidized to aluminum ion... $Al(s) rarr Al^(3+) +3e^-$ $(i)$ Hydrogen ion is REDUCED to dihydrogen... $H^+ +e^(-) rarr 1/2H_2(g)$ $(ii)$ We add $(i)+3xx(ii)$ $Al(s)+3H^+ +3e^(-) rarr Al^(3+)...
1 Answers 1 viewsAnd we could represent this by half-equations....CLEARLY iron metal is oxidized.... $FerarrFe^(3+) + 3e^(-)$ $(i)$ And meanwhile, the potent OXIDANT, dioxygen gas is reduced to $O^(2-)$... $O_2(g) + 4e^(-) rarr...
1 Answers 1 viewsAs per the equation , $2$ moles of Aluminum Oxide ($Al_2O_3$) produces $4$ moles of Aluminum ($Al$) and $3$ moles of Oxygen ($O_2$). The ratio of $Al$ to $O$ is...
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 views