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3 years ago

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The reduction of nitrogen monoxide is described by the following chemical equation: 2H2 (g) +2NO (g) 2H20 ()N2 (g Suppose a two-

step mechanism is proposed for this reaction, beginning with this elementary reaction: H2 g+2NO(g)- N20 (g)+H20(g) Suppose also that the second step of the mechanism should be bimolecular Suggest a reasonable second step. That is, write the balanced chemical equation of a bimolecular elementary reaction that would complete the proposed mechanism

1 answer:

Ipatiy [6.2K]3 years ago

8 0

Answer:

Reasonable Second step- $N_{2}O(g)+H_{2}(g)\rightarrow N_{2}(g)+H_{2}O(g)$

Explanation:

The given single step chemical reaction is as follows.

$2H_{2}(g)+2NO \rightarrow 2H_{2}O(g)+N_{2}(g)$

Suppose a two-step mechanism is proposed for this reaction,

The reaction occured in two steps they are as follows.

$Step -1:H_{2}(g)+2NO \rightarrow N_{2}O(g)+H_{2}O(g)$

$Step-2:N_{2}O(g)+H_{2}(g)\rightarrow N_{2}(g)+H_{2}O(g)$

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How many moles of MgCl2 are there in 319 g of the compound?

Leokris [45]

Hey there!:

Molar mass MgCl2 = 95.2110 g/mol

So:

1 mole MgCl2 -------------- 95.2110 g

moles MgCl2 ---------------- 319 g

moles MgCl2 = 319 * 1 / 95.2110

moles MgCl2 = 319 / 95.2110

=> 3.350 moles of MgCl2

Hope that helps!

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3 years ago

Nitroglycerin is a dangerous powerful explosive that violently decomposes when it is shaken or dropped. The Swedish chemist Alfr

Ganezh [65]

Answer:

a. $4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)$

b. 146.0 g

Explanation:

Question 1 (a). Just as the problem states, liquid nitroglycerin decomposes into nitrogen gas $N_2$ , oxygen gas $O_2$ , water vapor $H_2O$ and carbon dioxide $CO_2$ . Let's write the decomposition of nitroglycerin into these 4 components:

$C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + CO_2 (g)$

Now we need to balance the equation. Firstly, notice we have 3 carbon atoms on the left and 1 on the right, so let's multiply carbon dioxide by 3:

$C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)$

Now, we have 3 nitrogen atoms on the left and 2 on the right, so let's multiply nitrogen on the right by $\frac{3}{2}$ :

$C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)$

We have 5 hydrogen atoms on the left, 2 on the right, so let's multiply the right-hand side by $\frac{5}{2}$ :

$C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)$

Finally, count the oxygen atoms. We have a total of 9 on the left. On the right we have (excluding oxygen molecule):

$\frac{5}{2} + 6 = 8.5$

This leaves $9 - 8.5 = 0.5 = \frac{1}{2}$ of oxygen. Since oxygen is diatomic, we need to take one fourth of it to get one half in total:

$C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + \frac{1}{4} O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)$

To make it look neater without fractional coefficients, multiply both sides by 4:

$4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)$

Question 2 (b). Now we can make use of the balanced chemical equation and apply it for the context of this separate problem. We're given the following variables:

$V_{CO_2} = 41.0 L$

$T = -14.0^oC + 273.15 K = 259.15 K$

$p = 1 atm$

Firstly, we may find moles of carbon dioxide produced using the ideal gas law $pV = nRT$ .

Rearranging for moles, that is, dividing both sides by RT (here R is the ideal gas law constant):

$n_{CO_2} = \frac{pV_{CO_2}}{RT} = \frac{1 atm\cdot 41.0 L}{0.08206 \frac{L atm}{mol K}\cdot 259.15 K} = 1.928 mol$

According to the stoichiometry of the balanced chemical equation:

$4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)$

4 moles of nitroglycerin (ng) produce 12 moles of carbon dioxide. From here we can find moles o nitroglycerin knowing that:

$\frac{n_{ng}}{4} = \frac{n_{CO_2}}{12} \therefore n_{ng} = \frac{4}{12}n_{CO_2} = \frac{1}{3}\cdot 1.928 mol = 0.6427 mol$

Multiplying the number of moles of nitroglycerin by its molar mass will yield the mass of nitroglycerin decomposed:

$m_{ng} = n_{ng}\cdot M_{ng} = 0.6427 mol\cdot 227.09 g/mol = 146.0 g$

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