Question #a73d6

Dec 11, 2014

Let us look at the following equations

3N${O}_{2}$ + ${H}_{2}$O −→ 2HN${O}_{3}$ + NO.

As per the above equation 3 moles of N${O}_{2}$ are needed to form or get 2 moles of HN${O}_{3}$.

in terms of molar mass , one mole of N${O}_{2}$ has mass 46.0 g and one mole of HN${O}_{3}$ has mass 63.0 g ( approx.)

3 mole of N${O}_{2}$ produces 2 moles of HN${O}_{3}$

46 x 3 g of N${O}_{2}$ produces 2 x 63 g of HN${O}_{3}$

138 g of N${O}_{2}$ produces 126 g of HN${O}_{3}$

To get 1 g of HN${O}_{3}$ we will need ( 138g / 126g ) of N${O}_{2}$

To get 1 g of HN${O}_{3}$ we will need 1.10g of N${O}_{2}$

To get 63.0 g of HN${O}_{3}$ we will need 63 x 1.10 g of N${O}_{2}$

69.3 g of N${O}_{2}$

Dec 11, 2014

The answer is $83 g$.

Starting from the balanced chemical equation

$3 N {O}_{2} + {H}_{2} O \to 2 H N {O}_{3} + N O$

one can see that we have a $3 : 2$ mole ratio between $N {O}_{2}$ and ${H}_{2} O$; that is, for every $3$ moles of $N {O}_{2}$ that react, $2$ moles of $H N {O}_{3}$ are formed.

Now, starting from the given mass of nitric acid, ${m}_{H N {O}_{3}} = 75 g$, one can determine the number of moles formed in the reaction and, by working backwards, determine how many moles of $N {O}_{2}$ were used. So,

${n}_{H N {O}_{3}} = \frac{m a s {s}_{H N {O}_{3}}}{m o l a r m a s s} = \frac{75 g}{63 \frac{g}{m o l}} = 1.2$ moles (knowing that nitric acid's molar mass is $63 \frac{g}{m o l}$).

Therefore, the number of $N {O}_{2}$ moles used is

${n}_{N {O}_{2}} = \frac{3}{2} \cdot 1.2 = 1.8$ moles

Using the same conversion between moles and mass, one can determine that

${m}_{N {O}_{2}} = 1.8 m o l e s \cdot 46 \frac{g}{m o l} = 83 g$ ($N {O}_{2}$'s molar mass is $46 \frac{g}{m o l}$).