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3 years ago

How many moles of the product fecl3 do you make, when 8moles of fe react with excess cl2

2 answers:

7 0

To figure out questions related to reacting moles/masses, the first step is always to write a complete balanced equation.

2Fe (s) + 3Cl2 (g) → 2FeCl3 (s)

Since Cl2 is the excess reactant, and Fe is the limiting reactant, we can simply find the number of moles of the product by comparing the mole ratio of the limiting reactant to the mole ratio of the product from the equation.

From the equation, mole ratio of Fe:FeCl3 = 2:2 = 1:1, the number of moles of product is exactly the same as the number of moles of the limiting reactant, which makes it 8 moles.

Note that if the mole ratio is not 1:1, you have to do some calculations to make sure the no. of moles is balanced at the end. Which means, if the mole ratio happened to be 1:2, the no. of moles of the product would be 8x2=16 instead.

So, your answer is 8 moles.

Yuri [45]3 years ago

3 0

Hello There,

The answer is:

8moles

Hope this help have a great da:)

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A 0.2722 g sample of a pure carbonate, X n CO 3 ( s ) , was dissolved in 50.0 mL of 0.1200 M HCl ( aq ) . The excess HCl ( aq )

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Answer:

0.00369 moles of HCl react with carbonate.

Explanation:

Number of moles of HCl present initially = $\frac{0.1200}{1000}\times 50.0$ moles = 0.00600 moles

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According to above equation, 1 mol of NaOH reacts with 1 mol of 1 mol of HCl.

So, excess number of moles of HCl present = number of NaOH added for back titration = $\frac{0.0980}{1000}\times 23.60$ moles = 0.00231 moles

So, mole of HCl reacts with carbonate = (Number of moles of HCl present initially) - (excess number of moles of HCl present) = (0.00600 - 0.00231) moles = 0.00369 moles

Hence, 0.00369 moles of HCl react with carbonate.

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marysya [2.9K]

Answer:

The degree of dissociation of acetic acid is 0.08448.

The pH of the solution is 3.72.

Explanation:

The $pK_a=4.756$

The value of the dissociation constant = $K_a$

$pK_a=-\log[K_a]$

$K_a=10^{-4.756}=1.754\times 10^{-5}$

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Degree of dissociation = α

$HAc\rightleftharpoons H^++Ac^-$

Initially

At equilibrium ;

(c-cα) cα cα

The expression of dissociation constant is given as:

$K_a=\frac{[H^+][Ac^-]}{[HAc]}$

$1.754\times 10^{-5}=\frac{c\times \alpha \times c\times \alpha}{(c-c\alpha)}$

$1.754\times 10^{-5}=\frac{c\alpha ^2}{(1-\alpha)}$

$1.754\times 10^{-5}=\frac{0.00225 \alpha ^2}{(1-\alpha)}$

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α = 0.08448

The degree of dissociation of acetic acid is 0.08448.

$[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M$

The pH of the solution ;

$pH=-\log[H^+]$

$=-\log[0.0001901 M]=3.72$

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