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12345 [234]
2 years ago
6

N2 (g) + 3 H2 (g) → 2 NH3 (g)

Chemistry
1 answer:
shusha [124]2 years ago
3 0

This problem is asking for the rate of disappearance of gaseous nitrogen, given the rate of appearance of ammonia and the chemical reaction. At the end, the result turns out to be -0.228 M/s.

<h3>Rates of appearance and disappearance</h3>

In chemical kinetics, one of the most relevant calculations are based on rates of appearance and disappearance of chemical species in a chemical reaction. This can be calculated via rate portions based on the stoichiometric coefficients in the reaction.

Thus, for this problem, one can write:

\frac{r_{N_2}}{-1} =\frac{r_{H_2}}{-3} =\frac{r_{NH_3}}{2}

Where the rate of appearance or disappearance is divided by the stoichiometric coefficient. Therefore, one can solve for the rate of disappearance of N2 with:

\frac{r_{N_2}}{-1} =\frac{r_{NH_3}}{2}\\\\r_{N_2}=\frac{r_{NH_3}*-1}{2}\\\\r_{N_2}=\frac{0.456M/s*-1}{2}\\\\r_{N_2}=-0.228M/s

Learn more about chemical kinetics: brainly.com/question/26351746

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How many liters of 4.0 M NaOH solution will react with 1.2 mol H2SO4? (Remember to balance the equation.)
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The answer is D. 0.60 L

Explanation:

The balanced reaction equation including states of matter is;

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

More simple:

H2SO4 + 2NaOH → Na2SO4 + 2H2O

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Number of moles of H2SO4 reacted = 1.2 moles

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2 moles of NaOH reacts with 1 mole of H2SO4

x moles of NaOH reacts with 1.2 moles of H2SO4

x = 2 * 1.2/1 = 2.4 moles of NaOH

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Number of moles = Concentration * Volume

Volume = number of moles/concentration

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Volume = 2.4 moles/ 4.0 M

Volume =  0.60 L

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