What is the theoretical yield of aluminum oxide if 1.60 mol of aluminum metal is exposed to 1.50 mol of oxygen?

Ionic bonds occur between anions and cations. And cations may be metallic, i.e., $M^(2+)$, or they may be non-metallic, i.e. ammonium salts, $NH_4^+$. And anions, while typically non-metallic, i.e. $X^-$...

1 mole of aluminum oxide has a molecular weight of 102 grams (Al=27 grams and O=16 grams). It means that for 1000 kg of compound you can get 529 kilograms...

This is a limiting reactant problem. We know that we will need a balanced equation with masses, molar masses, and moles of the involved. 1. Gather all the information...

This is a double replacement (double displacement reaction). Balanced Equation $"CsCl(aq)"+"AgNO"_3("aq")$$rarr$$"CsNO"_3("aq")+"AgCl(s)"$ Molar Masses We need the molar masses of $"CsCl"$ and $"AgCl"$. $"CsCl:"$$"168.358452 g/mol"$ $"AgCl:"$$"143.3212 g/mol"$ Theoretical...

$2NaCl + CaO rarr CaCl_2 + Na_2O$ $"Moles of calcium oxide"$ $=$ $(20*g)/(56.08*g*mol^-1)=0.357*mol.$ Given the stoichiometric equation, which you have kindly provided, clearly, at most, I can get an equivalent...

Let's define those three terms before we get started: The percent yield of a reaction is the ratio of actual yield to the theoretical yield times one-hundred percent....

And thus for this example $%"Yield"$ $=$ $(7.2*g)/(10*g)xx100%$ And we don't even need a calculator......... Most of the time, we write a stoichiometric equation, that shows the molar equivalence between...

Actual yield in a reaction is almost always less than the theoretical yield, primarily because losses of the substances involved may occur anywhere in an experiment. Otherwise, there can be...

A reaction's is simply the number of grams of a product that you actually get for every $"100 g"$ of said product that you could theoretically get. The...

The actual yield is a term used to describe the amount of product that is actually produced by a given chemical reaction. By comparison, the theoretical yield is a...