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Archy [21]
3 years ago
10

Determine the limiting reactant in a mixture containing 139 g of Se, 431 g of Cl2, and 110 g of H2O. Calculate the maximum mass

(in grams) of selenic acid, H2SeO4, that can be produced in the reaction.
Chemistry
1 answer:
Bad White [126]3 years ago
3 0

Answer : The limiting reagent is H_2O

The mass of H_2SeO_4 produced is, 220.4 grams.

Explanation : Given,

Mass of Se = 139 g

Mass of Cl_2 = 431 g

Mass of H_2O = 110 g

Molar mass of Se = 79 g/mol

Molar mass of Cl_2 = 71 g/mol

Molar mass of H_2O = 18 g/mol

First we have to calculate the moles of Se, Cl_2 and H_2O.

\text{Moles of }Se=\frac{\text{Given mass }Se}{\text{Molar mass }Se}

\text{Moles of }Se=\frac{139g}{79g/mol}=1.76mol

and,

\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}

\text{Moles of }Cl_2=\frac{431g}{71g/mol}=6.07mol

and,

\text{Moles of }H_2O=\frac{\text{Given mass }H_2O}{\text{Molar mass }H_2O}

\text{Moles of }H_2O=\frac{110g}{18g/mol}=6.11mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

Se(s)+3Cl_2(g)+4H_2O(l)\rightarrow H_2SeO_4(aq)+6HCl(aq)

From the balanced reaction we conclude that:

As, 4 mole of H_2O react with 1 mole of Se

So, 6.11 moles of HCl react with \frac{6.11}{4}=1.52 moles of Se

and,

As, 4 mole of H_2O react with 3 mole of Cl_2

So, 6.11 moles of HCl react with \frac{6.11}{4}\times 3=4.58 moles of Cl_2

From this we conclude that, Cl_2\text{ and }Se are excess reagent because the given moles are greater than the required moles and H_2O is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of H_2SeO_4

From the reaction, we conclude that

As, 4 mole of H_2O react to give 1 mole of H_2SeO_4

So, 6.11 mole of H_2O react to give \frac{6.11}{4}=1.52 mole of H_2SeO_4

Now we have to calculate the mass of H_2SeO_4

\text{ Mass of }H_2SeO_4=\text{ Moles of }H_2SeO_4\times \text{ Molar mass of }H_2SeO_4

Molar mass of H_2SeO_4 = 145 g/mole

\text{ Mass of }H_2SeO_4=(1.52moles)\times (145g/mole)=220.4g

Therefore, the mass of H_2SeO_4 produced is, 220.4 grams.

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Answer:

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Explanation:

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3Mg(s) +2P(s) -------> Mg3P2(s) + energy

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A sample of gas has an initial pressure of 1.5 atm, an initial volume of 3.0 L, and an initial temperature of 293K. If the final
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Explanation:

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Step 2: Calculate the final volume of the gas

If we assume ideal behavior, we can calculate the final volume of the gas using the combined gas law.

P₁ × V₁ / T₁ = P₂ × V₂ / T₂

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<u>Answer:</u> The energy of one photon of the given light is 3.79\times 10^{-19}J

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