Question

how many grams of $\ce{nbr_{3}}$ can be made from $2.22\times10^{-2}$ moles of $\ce{br_{2}}$? (answer = $3.76\times10^{0}$ g)
$\ce{n_{2} + 3br_{2}\to2nbr_{3} + 44.0kj}$
the molar mass of $\ce{nbr_{3}}$ is 1 g
2 mol $\ce{nbr_{3}}$
$(2.22\times10^{2}\text{ mol }\ce{br_{2}})(\dfrac{}{}) = $3 mol $\ce{nbr_{3}}$
4 mol $\ce{br_{2}}$
5 g $\ce{nbr_{3}}$
6 mol $\ce{nbr_{3}}$$(\dfrac{}{}) = $__7__ g $\ce{nbr_{3}}$
8 mol $\ce{nbr_{3}}$
a. 1 b. 2 c. 3 d. 4 e. 5 f. 253.719 g. $3.76\times10^{0}$
h. $3.26\times10^{3}$ i. 44.0 j. $1.48\times10^{-2}$ k. $9.77\times10^{3}$ l. $4.44\times10^{2}$
m. $6.66\times10^{2}$ 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 molar mass of \(NBr_3\)

The molar mass of \(N\) is \(14.007\space g/mol\) and of \(Br\) is \(79.904\space g/mol\).
For \(NBr_3\), \(M = 14.007+3\times79.904=14.007 + 239.712=253.719\space g/mol\)

Step2: Use mole ratio from balanced equation

From \(N_2 + 3Br_2
ightarrow2NBr_3\), the mole ratio of \(NBr_3\) to \(Br_2\) is \(\frac{2}{3}\)
Moles of \(NBr_3=(2.22\times 10^{- 2}\space mol\space Br_2)\times\frac{2\space mol\space NBr_3}{3\space mol\space Br_2}=1.48\times 10^{-2}\space mol\space NBr_3\)

Step3: Calculate mass of \(NBr_3\)

Mass \(m=n\times M\), where \(n = 1.48\times 10^{-2}\space mol\) and \(M = 253.719\space g/mol\)
\(m=(1.48\times 10^{-2}\space mol)\times253.719\space g/mol\approx3.76\space g\)

Answer:

1. F. \(253.719\)
2. B. \(2\)
3. J. \(1.48\times 10^{-2}\)
4. D. \(3\)
5. F. \(253.719\)
6. J. \(1.48\times 10^{-2}\)
7. G. \(3.76\times 10^{0}\)
8. B. \(1\)