Question

In Ostwald's process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.0 g of ammonia and 20.0 g of oxygen?

Answer 1

"The balanced chemical equation for the given reaction is given as:

$\begin{matrix} 4{ NH }_{ 3 }(g)\quad + \\ 4\times17g \\ =\quad 68g \end{matrix}\begin{matrix} 5{ O }_{ 2 }\quad (g)\quad\rightarrow \\ 2\times32g \\ =\quad 160g \end{matrix}\begin{matrix}4NO(g)\quad + \\ 4\times30g \\ =\quad 120g \end{matrix}\begin{matrix} 6{ H }_{ 2 }O(g) \\ 6\times18g \\ =\quad 108g \end{matrix}$

Thus, 68 g of${ NH }_{ 3 }$ reacts with 160 g of ${ O }_{ 2 }$.

Therefore, 10 g of${ NH }_{ 3 }$ reacts with$\frac { 160\quad \times \quad 10 }{ 68 } \quad g of { O }_{ 2 }, or\quad 23.53 g of { O }_{ 2 }.$

But the available amount of ${ O }_{ 2 }$ is 20 g.

Therefore,${ O }_{ 2 }$is the "limiting reagent" (we have considered the amount of ${ O }_{ 2 }$to calculate the "weight of nitric oxide"obtained in the "reaction").

Now, 160 g of${ O }_{ 2 }$ gives 120 g of NO.

Therefore, 20 g of ${ O }_{ 2 }$gives $\frac { 120\quad \times \quad 20 }{ 160 } \quad g$ of N, or 15 g of NO.

Hence, a "maximum" of "15 g" of "nitric oxide" can be obtained."

Answer 2

Risk capital establishment established for:-

a) Import analysis
b) Export analysis
c) none
d) both