Question

missed this? watch kcv limiting reactant, theoretical yield and percent yield. we limiting reactant and theoretical yield. read section 4.4. you can click on the review link to access the section in your etext. find the limiting reactant for each initial amount of reactants 2na(s) + br₂(g) → 2nabr(s) part c 1.5 mol na and 2.1 mol br₂ express your answer as a chemical formula. view available hint(s) part d 12.4 mol na and 5.5 mol br₂ express your answer as a chemical formula. view available hint(s)

Explanation:

Step1: Identify the mole - ratio from the balanced equation

The balanced chemical equation is \(2Na(s)+Br_2(g)
ightarrow 2NaBr(s)\). The mole - ratio of \(Na\) to \(Br_2\) is \(n_{Na}:n_{Br_2}=2:1\).

Step2: Calculate the amount of \(Br_2\) required to react with \(Na\) (for Part C)

Given \(n_{Na} = 1.5\ mol\). According to the mole - ratio, the amount of \(Br_2\) required to react completely with \(Na\) is \(n_{Br_2}^{required}=\frac{1}{2}n_{Na}\). Substituting \(n_{Na}=1.5\ mol\), we get \(n_{Br_2}^{required}=\frac{1.5}{2}=0.75\ mol\). Since we have \(n_{Br_2}=2.1\ mol\) and \(0.75\ mol<2.1\ mol\), \(Na\) is the limiting reactant.

Step3: Calculate the amount of \(Na\) required to react with \(Br_2\) (for Part D)

Given \(n_{Br_2}=5.5\ mol\). According to the mole - ratio, the amount of \(Na\) required to react completely with \(Br_2\) is \(n_{Na}^{required}=2n_{Br_2}\). Substituting \(n_{Br_2}=5.5\ mol\), we get \(n_{Na}^{required}=2\times5.5 = 11\ mol\). Since we have \(n_{Na}=12.4\ mol\) and \(11\ mol<12.4\ mol\), \(Br_2\) is the limiting reactant.

Answer:

Part C: \(Na\)
Part D: \(Br_2\)