Question #a73d6

Question

Question #a73d6

2 Answers

Impact of this question

2670 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-12-11T05:17:01

Let us look at the following equations

3N $O_{2}$ $O_2$ + $H_{2}$ $H_2$ O −→ 2HN $O_{3}$ $O_3$ + NO.

As per the above equation 3 moles of N $O_{2}$ $O_2$ are needed to form or get 2 moles of HN $O_{3}$ $O_3$ .

in terms of molar mass , one mole of N $O_{2}$ $O_2$ has mass 46.0 g and one mole of HN $O_{3}$ $O_3$ has mass 63.0 g ( approx.)

3 mole of N $O_{2}$ $O_2$ produces 2 moles of HN $O_{3}$ $O_3$

46 x 3 g of N $O_{2}$ $O_2$ produces 2 x 63 g of HN $O_{3}$ $O_3$

138 g of N $O_{2}$ $O_2$ produces 126 g of HN $O_{3}$ $O_3$

To get 1 g of HN $O_{3}$ $O_3$ we will need ( 138g / 126g ) of N $O_{2}$ $O_2$

To get 1 g of HN $O_{3}$ $O_3$ we will need 1.10g of N $O_{2}$ $O_2$

To get 63.0 g of HN $O_{3}$ $O_3$ we will need 63 x 1.10 g of N $O_{2}$ $O_2$

69.3 g of N $O_{2}$ $O_2$

Answer 2 · 2014-12-11T08:58:38

The answer is $83 g$ $83g$ .

Starting from the balanced chemical equation

$3 N O_{2} + H_{2} O \to 2 H N O_{3} + N O$ $3NO_2 + H_2O -> 2HNO_3 + NO$

one can see that we have a $3 : 2$ $3:2$ mole ratio between $N O_{2}$ $NO_2$ and $H_{2} O$ $H_2O$ ; that is, for every $3$ $3$ moles of $N O_{2}$ $NO_2$ that react, $2$ $2$ moles of $H N O_{3}$ $HNO_3$ are formed.

Now, starting from the given mass of nitric acid, $m_{H N O_{3}} = 75 g$ $m_(HNO_3) = 75g$ , one can determine the number of moles formed in the reaction and, by working backwards, determine how many moles of $N O_{2}$ $NO_2$ were used. So,

$n_{H N O_{3}} = \frac{m a s s_{H N O_{3}}}{m o l a r m a s s} = \frac{75 g}{63 \frac{g}{m o l}} = 1.2$ $n_(HNO_3) = (mass_(HNO_3))/(molarmass) = (75g)/(63 g/(mol)) = 1.2$ moles (knowing that nitric acid's molar mass is $63 \frac{g}{m o l}$ $63g/(mol)$ ).

Therefore, the number of $N O_{2}$ $NO_2$ moles used is

$n_{N O_{2}} = \frac{3}{2} \cdot 1.2 = 1.8$ $n_(NO_2) = 3/2 * 1.2 = 1.8$ moles

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

$m_{N O_{2}} = 1.8 m o l e s \cdot 46 \frac{g}{m o l} = 83 g$ $m_(NO_2) = 1.8 mol es * 46 g/(mol) = 83g$ ( $N O_{2}$ $NO_2$ 's molar mass is $46 \frac{g}{m o l}$ $46 g/(mol)$ ).

Science

Math

Humanities

... and beyond

Question #a73d6

2 Answers

Related questions

Impact of this question