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LuckyWell [14K]
2 years ago
7

The psychoactive drug sold as a methamphetamine - "speed" (c10h15n) - undergoes a series of reactions in the body; the net resul

t of these reactions is the oxidation of solid methamphetamine by oxygen gas to produce carbon dioxide gas, liquid water, and nitrogen gas. what is the coefficient of nitrogen in the balanced equation for the reaction? (balance the equation with the smallest possible whole number coefficients.)
Chemistry
1 answer:
Phoenix [80]2 years ago
8 0

Answer: The coefficient of nitrogen in the given equation is 2.

Explanation: The reaction for the oxidation of methamphentamine with oxygen gas in the body is given by:

4C_{10}H_{15}N(s)+55O_2(g)\rightarrow 40CO_2(g)+30H_2O(l)+2N_2(g)

By Stoichiometry,

4 moles of methamphentamine reacts with 55 moles of oxygen gas to produce 40 moles of carbon dioxide gas, 30 moles of water and 2 moles of nitrogen gas.

Coefficient of C_{10}H_{15}N=4

Coefficient of O_2=55

Coefficient of CO_2=40

Coefficient of H_2O=30

Coefficient of N_2=2

Hence, the coefficient of nitrogen in the given equation is 2.

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Calculate δ h for the reaction:no (g) + o2 (g) ↔ no2 (g). given: 2o3(g) ↔ 3o2(g) δh=-426 kj o2(g) ↔ 2o(g) δh=+ 490 kj no(g) + o3
maks197457 [2]
To calculate the <span>δ h, we must balance first the reaction: 

NO + 0.5O2 -----> NO2

Then we write all the reactions,

2O3 -----> 3O2    </span><span>δ h = -426 kj        eq. (1)

O2 -----> 2O    </span><span>δ h = 490 kj             eq. (2)

NO + O3 -----> NO2 + O2    </span><span>δ h = -200 kj          eq. (3)


We divide eq. (1) by 2, we get

</span>O3 -----> 1.5O2    δ h = -213  kj             eq. (4)

Then, we subtract eq. (3) by eq. (4) 

NO + O3 ----->  NO2 + O2   δ h = -200 kj
-       (O3 -----> 1.5 O2         δ h = -213  kj)
NO -----> NO2 - 0.5O2        δ h = 13  kj               eq. (5)


eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)

O -----> 0.5O2  <span>δ h = -245  kj         eq. (6)
</span>
Add eq. (6) to eq. (5), we get

NO -----> NO2 - 0.5O2        δ h = 13  kj 
+  O -----> 0.5O2                 δ h = -245  kj
NO + O ----> NO2               δ h = -232 kj

<em>ANSWER:</em> <em>NO + O ----> NO2               δ h = -232 kj</em>


4 0
3 years ago
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