Question

1. if 60.0 g of barium and 5.00 g of phosphorus are available to react, which is the limiting reagent?

phosphorus
barium
phosphorus and barium
not enough information

Explanation:

Step1: Write the balanced chemical equation

The reaction between barium ($Ba$) and phosphorus ($P$) forms barium phosphide ($Ba_3P_2$). The balanced equation is $3Ba + 2P=Ba_3P_2$.

Step2: Calculate the molar - mass of reactants

The molar mass of $Ba$ is approximately $M_{Ba}=137.33\ g/mol$, and the molar mass of $P$ is approximately $M_{P}=30.97\ g/mol$.

Step3: Calculate the number of moles of each reactant

The number of moles of $Ba$, $n_{Ba}=\frac{m_{Ba}}{M_{Ba}}=\frac{60.0\ g}{137.33\ g/mol}\approx0.437\ mol$.
The number of moles of $P$, $n_{P}=\frac{m_{P}}{M_{P}}=\frac{5.00\ g}{30.97\ g/mol}\approx0.161\ mol$.

Step4: Determine the mole - ratio from the balanced equation

From the balanced equation $3Ba + 2P = Ba_3P_2$, the mole - ratio of $Ba$ to $P$ is $\frac{n_{Ba}}{n_{P}}=\frac{3}{2}=1.5$.

Step5: Compare the actual mole - ratio with the stoichiometric mole - ratio

If all $0.161\ mol$ of $P$ reacts, the moles of $Ba$ required, $n_{Ba\ required}=0.161\ mol\times\frac{3}{2}=0.2415\ mol$. But we have $0.437\ mol$ of $Ba$. So, $P$ is the limiting reagent.

Answer:

phosphorus