Question

Nitric acid is produced commercially by the Ostwald process, represented by the following equations.

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
2 NO(g) + O2(g) 2 NO2(g)
3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g)
What mass in kg of NH3 must be used to produce 5.3 multiplied by 106 kg HNO3 by the Ostwald process, assuming 100% yield in each reaction?

Answer 1

5.3 multiplied by 106 kg HNO3 ?

so 561.8 kg?

Answer 2

no 5.3 * 10^6

Answer 3

ohh okay. we'll add up the reactions into 1 to make it easier.

then find how many moles of HNO3 there are in 5.3 * 10^6 kg

from here you wanna build a ratio:

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For a general reaction: \(\color{red}{a}A + \color{blue}{b}B\) \(\rightleftharpoons\) \( \color{green}{c}C\)

where upper case are the species (A,B,C), and lower case (a,b,c) are the coefficients,

\(\dfrac{n_A}{\color{red}{a}}=\dfrac{n_B}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\)

from here, you can solve for moles then convert to grams.

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in your case you would have:

\(\dfrac{n_{HNO_3}}{\color{red}{2}}=\dfrac{n_{NH_3}}{\color{blue}{4}}\)

find the moles of NH3. To convert mass to moles, use the relationship:

\(n=\dfrac{m}{M}\)

where, M=molar mass, m=mass, and n= moles.

don't forget to express your answer in kg as they asked.

Answer 4

i got 25.5506 g
do i change that to kg now

Answer 5

the answer isn't 25 g. you started with like a million kilograms