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marta [7]
2 years ago
13

Please balance the equation C7H16 + CoF3 = C7F16 + HF +CoF2 with solutions

Chemistry
1 answer:
Andrei [34K]2 years ago
3 0

Answer:

C₇H₁₆ + 32CoF₃ —> C₇F₁₆ + 16HF + 32CoF₂

Explanation:

C₇H₁₆ + CoF₃ —> C₇F₁₆ + HF + CoF₂

The above equation can be balance as illustrated below:

C₇H₁₆ + CoF₃ —> C₇F₁₆ + HF + CoF₂

There are 16 atoms of H on the left side and 1 atom on the right side. It can be balance by writing 16 before HF as shown below:

C₇H₁₆ + CoF₃ —> C₇F₁₆ + 16HF + CoF₂

There are 3 atoms of F on the left side and a total of 34 atoms on the right side. It can be balance by writing 32 before CoF₃ and 32 before CoF₂ as shown below:

C₇H₁₆ + 32CoF₃ —> C₇F₁₆ + 16HF + 32CoF₂

Now, the equation is balanced.

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A 10.5 mL sample of vinegar, containing acetic acid, was titrated using 0.460 M NaOH solution. The titration required 19.13 mL o
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Explanation:

Step 1:

A good first step for a problem like this is to write down the chemical formula and balance it.

It appears here that we have 10.5 mL of vinegar, which IS acetic acid, and 19.13 mL of 0.460 M NaOH. That will give us the following balanced chemical equation:

CH3COOH + NaOH ------> NaCH3COO + H2O

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Step 2:

Since all of our stoichiometric coefficients are one, we can use a shortcut to answer this equation. I don't know if it has a name, but I just call it the titration formula. It goes something like this:

M1 * V1 = M2 * V2

M stands for Molarity and V stands for volume. 1 and 2 being the before the reaction and after the reaction.

So, our M1 for this is going to be what the question says was used for this titration. That's 0.460M NaOH.

Our V1 is going to be the initial volume of the sample, which was 10.5 mL

Our V2 is going to be 19.13, which is the volume when we're finished.

It's clear that we don't know M2, so let's find it.

Keep in mind that it's easier to convert to liters pretty much always, so I've done that by dividing the mL values each by 1000.

Using some algebra, we can see that we now have:

0.460 M * 0.0105 L = x M * 0.01913 L

Which goes to:

\frac{0.00483mol}{0.01913L} = 0.252 M

<h3>So our M2, the molar concentration of acetic acid in this vinegar, is equal to 0.252 M. </h3>
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