Question #eb29e

Feb 27, 2017

$1.87 \cdot {10}^{23}$ formula units of $B a {\left(N {O}_{3}\right)}_{2}$

Explanation:

First compute the molar mass of $B a {\left(N {O}_{3}\right)}_{2}$:

MM $= 137.3 g + \left(2 \cdot 14.007 g\right) + \left(6 \cdot 15.999 g\right) = 261.3 \frac{g B a {\left(N {O}_{3}\right)}_{2}}{m o l}$

Next use dimensional analysis to calculate the answer. Be sure to set up the units correctly and use appropriate conversions. Usually if you set up the units correctly, you'll get the correct answer.

$\left(81.3 \cancel{g B a {\left(N {O}_{3}\right)}_{2}}\right) \left(\frac{1 \cancel{m o l B a {\left(N {O}_{3}\right)}_{2}}}{261.3 \cancel{g B a {\left(N {O}_{3}\right)}_{2}}}\right) \left(\frac{6.02 \cdot {10}^{23} F U}{1 \cancel{m o l B a {\left(N {O}_{3}\right)}_{2}}}\right) = 1.87 \cdot {10}^{23} F U$