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Solve for [tex]\( x \)[/tex]:

[tex]\[ 3x = 6x - 2 \][/tex]



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[tex]$
2 C_2H_2(g) + 5 O_2(g) \rightarrow 4 CO_2(g) + 2 H_2O(g)
$[/tex]

How many liters of [tex]\(C_2H_2\)[/tex] are required to produce [tex]\(8 \, \text{L}\)[/tex] of [tex]\(CO_2\)[/tex], assuming the reaction is at STP?

The ratios are: [tex]\(\frac{2 \, \text{L} \, C_2H_2}{4 \, \text{L} \, CO_2}\)[/tex] or [tex]\(\frac{4 \, \text{L} \, CO_2}{2 \, \text{L} \, C_2H_2}\)[/tex]

[tex]\[ \text{[?]} \, \text{L} \, C_2H_2 \][/tex]



Answer :

GinnyAnswer
Alright, let's solve this step-by-step:

We have the balanced chemical equation:
[tex]\[ 2 \, \text{C}_2\text{H}_2 (\text{g}) + 5 \, \text{O}_2 (\text{g}) \rightarrow 4 \, \text{CO}_2 (\text{g}) + 2 \, \text{H}_2\text{O} (\text{g}) \][/tex]

We need to determine how many liters of [tex]\(\text{C}_2\text{H}_2\)[/tex] are required to produce 8 liters of [tex]\(\text{CO}_2\)[/tex].

From the balanced equation, we know the volume ratios are as follows:
[tex]\[ 2 \, \text{L} \, \text{C}_2\text{H}_2 \text{ produces } 4 \, \text{L} \, \text{CO}_2 \][/tex]

We can set up the ratio of [tex]\(\text{C}_2\text{H}_2\)[/tex] to [tex]\(\text{CO}_2\)[/tex] from the equation:
[tex]\[ \frac{2 \, \text{L} \, \text{C}_2\text{H}_2}{4 \, \text{L} \, \text{CO}_2} \][/tex]

Let [tex]\( x \)[/tex] be the volume of [tex]\(\text{C}_2\text{H}_2\)[/tex] needed to produce 8 liters of [tex]\(\text{CO}_2\)[/tex]:

[tex]\[ \frac{2 \, \text{L} \, \text{C}_2\text{H}_2}{4 \, \text{L} \, \text{CO}_2} = \frac{x \, \text{L} \, \text{C}_2\text{H}_2}{8 \, \text{L} \, \text{CO}_2} \][/tex]

To solve for [tex]\( x \)[/tex], we can cross-multiply:

[tex]\[ 2 \times 8 = 4 \times x \][/tex]
[tex]\[ 16 = 4x \][/tex]

Now, we divide both sides by 4 to isolate [tex]\( x \)[/tex]:

[tex]\[ x = \frac{16}{4} \][/tex]
[tex]\[ x = 4 \][/tex]

Therefore, the volume of [tex]\(\text{C}_2\text{H}_2\)[/tex] required to produce 8 liters of [tex]\(\text{CO}_2\)[/tex] is [tex]\( 4 \)[/tex] liters.

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