Quantitative Chemistry - Reacting Masses and Limiting Reactants

Calculate the mass of Magnesium Oxide ($MgO$) produced when $48\text{ g}$ of Magnesium ($Mg$) burns completely in oxygen.

1. Write the balanced equation: $2Mg + O_2 \rightarrow 2MgO$
2. Calculate the moles of $Mg$ ($A_r$ of $Mg = 24$): $n(Mg) = \frac{48}{24} = 2\text{ mol}$
3. Use the molar ratio from the equation ($2:2$ or $1:1$): $n(MgO) = n(Mg) = 2\text{ mol}$
4. Calculate the mass of $MgO$ ($M_r$ of $MgO = 24 + 16 = 40$): $m(MgO) = 2 \times 40 = 80\text{ g}$

First, convert the given mass to moles. Then, use the stoichiometric coefficients from the balanced equation to find the moles of the product. Finally, convert those moles back into mass.

If $10\text{ g}$ of Hydrogen ($H_2$) reacts with $64\text{ g}$ of Oxygen ($O_2$) to form water, which is the limiting reactant?

1. Balanced equation: $2H_2 + O_2 \rightarrow 2H_2O$
2. Calculate moles of reactants ($M_r$ of $H_2 = 2$, $M_r$ of $O_2 = 32$): $n(H_2) = \frac{10}{2} = 5\text{ mol}$ $n(O_2) = \frac{64}{32} = 2\text{ mol}$
3. Determine the required ratio: The equation requires $2$ moles of $H_2$ for every $1$ mole of $O_2$. For $2\text{ mol}$ of $O_2$, we need: $2 \times 2 = 4\text{ mol of } H_2$
4. Compare available vs. required: We have $5\text{ mol}$ of $H_2$, but only $4\text{ mol}$ is needed. Therefore, $O_2$ is the limiting reactant.

The limiting reactant is found by comparing the actual number of moles available to the stoichiometric requirements of the balanced equation. Since $O_2$ will run out first, it is the limiting reactant.