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PSYCHO15rus [73]
3 years ago
5

Determine the mole ratio between oxygen and water for the following equation:

Chemistry
1 answer:
vodka [1.7K]3 years ago
5 0

Answer:2.2

Explanation:

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Identify the combustion reaction. c4h12 + 7o2 ⟶ 6h2o + 4co2 2h2 + o2 ⟶ 2h2o al2s3 ⟶ 2al + 3s cl2 + 2kbr ⟶ 2kcl + br2
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From the given equations, the combustion reaction is;
C₄H₁₂ + 7O₂ --> 4CO₂ + 6H₂O
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Calcule a variação da entalpia dessa reação ( 2 NH3 (g) ---> CO(NH2)2 (s) + H2O (L) ) a partir das seguintes equações termoqu
Nitella [24]

ΔH = +438 kJ  

We have three equations:  

(I) N₂ + 3H₂ → 2NH₃; Δ<em>H</em> = -92 kJ  

(II) H₂ +½O₂ → H₂O; Δ<em>H</em> = -286 kJ  

(III) CO(NH₂)₂ + ³/₂O₂ → CO₂ + 2H₂O + N₂; Δ<em>H</em> = -632 kJ  

From these, we must devise the target equation:  

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; Δ<em>H</em> = ?  

_________________________________

The target equation has 2NH₃ on the left, so you <em>reverse equation (I)</em>.  

When you reverse an equation, you <em>reverse the sign of its ΔH</em>.  

(V) 2NH₃ → N₂ + 3H₂; Δ<em>H</em> = +92 kJ  

Equation (V) has 1N₂ on the right, and that is not in the target equation.  

You need an equation with 1N₂ on the left.  

<em>Reverse Equation (III).</em>  

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; Δ<em>H</em> = +632 kJ  

Equation <em>(VI)</em> has ³/₂O₂ on the right, and that is not in the target equation.  

You need ³/₂O₂ on the left.  

Multiply <em>Equation (II) by three</em>.  

When you multiply an equation by three, you <em>multiply its ΔH by thre</em>e.

(VII) 3H₂ +³/₂O₂ → 3H₂O; Δ<em>H</em> = -286 kJ  

Now, you add equations (V), (VI), and (VII), <em>cancelling species</em> that appear on opposite sides of the reaction arrows.  

When you add equations, you add their Δ<em>H</em> values.  

_______________________________________

We get the target equation (IV):  

(V) 2NH₃ → <u>N</u>₂ + <u>3H</u>₂;                                    ΔH = +  92 kJ  

(VI) CO₂ + <u>2H</u>₂<u>O</u> + <u>N</u>₂ → CO(NH₂)₂ + ³/₂<u>O</u>₂; ΔH = +632 kJ  

(VII) <u>3H</u>₂ +³/₂<u>O</u>₂ → <u>3</u>H₂O;                             ΔH =   -286 kJ

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O;          ΔH =  +438 kJ  


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