Quick questions on Mole calculations and reacting masses explained: O-Level Chemistry

Almost every reacting-mass question follows the same three steps:

When the amounts of two reactants are both given, one usually runs out first; this is the limiting reagent, and it controls how much product forms. The other reactant is in excess. To find which is limiting:

The theoretical yield is the mass (or moles) of product the equation predicts. The actual yield is what is really obtained, which is usually less, because reactions may not finish, products are lost in handling, or side reactions occur. The percentage yield compares the two:

State the three steps of a reacting-mass calculation. [1 mark]

$0.20$ mol of a gas is collected at r.t.p. Calculate its volume. (Molar gas volume $= 24.0\ \text{dm}^3$.)

$4.8\ \text{g}$ of magnesium ($A_r = 24$) and $4.8\ \text{g}$ of oxygen ($\text{O}_2$, $M_r = 32$) react by $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$. Identify the limiting reagent. [3 marks]