Question

question 10 (1 point)
calculate the % yield for problem number 9 if only 500.0 g of nbr₃ are actually made from 525 g br₂
n₂ + 3 br₂ → 2 nbr₃ + 44.0 kj
actual yield
% yield = ----------------- 100 = ----------------- 100 = 2 %
theoretical yield
a. 1 b. 2 c. 3 d. 4 e. 5 f. 253.719 g. 3.38 x 10⁰
h. 3.26 x 10³ i. 44.0 j. 1.33 x 10⁻² k. 9.77 x 10³ l. 4.44 x 10²
m. 6.66 x 10² n. 500.0 o. 556 p. 90.0 q. 3.2852 r. 159.81
s. 28.013 t. 1.750 u. 0.8750 v. 13.0 w. 0.46306 x. 1.3892
y. 24.5 z. 38.5 aa. 2.190 bb. evolved cc. absorbed
dd. exothermic ee. endothermic

Explanation:

Step1: Calculate the theoretical yield of \(NBr_3\)

First, find the molar mass of \(Br_2\) (\(M_{Br_2}=2\times79.904 = 159.808\space g/mol\)) and \(NBr_3\) (\(M_{NBr_3}=14.007+3\times79.904=253.719\space g/mol\))
The moles of \(Br_2\) used: \(n_{Br_2}=\frac{m_{Br_2}}{M_{Br_2}}=\frac{525\space g}{159.808\space g/mol}\approx3.2852\space mol\)
From the balanced equation \(N_2 + 3Br_2
ightarrow2NBr_3\), the mole ratio of \(Br_2\) to \(NBr_3\) is \(3:2\)
The moles of \(NBr_3\) produced (theoretical): \(n_{NBr_3}=\frac{2}{3}n_{Br_2}=\frac{2}{3}\times3.2852\space mol\approx2.190\space mol\)
The theoretical mass of \(NBr_3\): \(m_{NBr_3}^{theo}=n_{NBr_3}\times M_{NBr_3}=2.190\space mol\times253.719\space g/mol\approx556\space g\)

Step2: Calculate the percentage yield

The percentage yield formula is \(\%yield=\frac{actual\space yield}{theoretical\space yield}\times100\)
Given \(actual\space yield = 500.0\space g\) and \(theoretical\space yield=556\space g\)
\(\%yield=\frac{500.0}{556}\times 100\approx90.0\%\)

Answer:

1. N. \(500.0\)
2. P. \(90.0\)
3. O. \(556\)