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1 year ago

Identify the limiting reactant when 32. 0 g hydrogen is allowed to react with 16. 0 g oxygen

Chemistry

1 answer:

Mkey [24]1 year ago

3 0

Answer:

Oxygen is limiting reactant

Explanation:

2 H2 + O2 ======> 2 H2 O

from this equation (and periodic table) you can see that

4 gm of H combine with 32 gm O2

H / O = 4/32 = 1/8

32 /16 = 2/1 shows O is limiter

for 32 gm H you will need 256 gm O and you only have 16 gm

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3 years ago

Iron (III) oxide and hydrogen react to form iron and water, like this: Fe 03(s)+3H9)2Fe(s)+3HO) At a certain temperature, a chem

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The question is incomplete, here is the complete question:

Iron (III) oxide and hydrogen react to form iron and water, like this:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

At a certain temperature, a chemist finds that a 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.

Compound Amount

Fe₂O₃ 3.95 g

H₂ 4.77 g

Fe 4.38 g

H₂O 2.00 g

Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

Answer: The value of equilibrium constant for given equation is $1.0\times 10^{-4}$

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

For hydrogen gas:

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

$\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M$

For water:

Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

$\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M$

For the given chemical equation:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

The expression of equilibrium constant for above equation follows:

$K_{eq}=\frac{[H_2O]^3}{[H_2]^3}$

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

$K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}$

Hence, the value of equilibrium constant for given equation is $1.0\times 10^{-4}$

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3 years ago

Based on the six given solvents, which of the following answers lists ALL of the solvents that are NOT suitable for liquid-liqui

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Answer:

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2 years ago

Which of the following statements explain why the van der Waals equation must be used to describe real gases? X. interactions be

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Answer:

Statements Y and Z.

Explanation:

The Van der Waals equation is the next one:

$nRT = (P + \frac{an^{2}}{V^{2}})(V -nb)$ (1)

The ideal gas law is the following:

$nRT = PV$ (2)

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As we can see from equation (1), the Van der Waals equation introduces two terms that correct the P and the V of the ideal gas equation (2), by the incorporation of the intermolecular interaction between the gases and the gases volume. The term an²/V² corrects the P of the ideal gas equation since the measured pressure is decreased by the attraction forces between the gases. The term nb corrects the V of the ideal gas equation, taking into account the volume occuppied by the gas in the total volume, which implies a reduction of the total space available for the gas molecules.

So, the correct statements are the Y and Z: the non-zero volumes of the gas particles effectively decrease the amount of "empty space" between them and the molecular attractions between gas particles decrease the pressure exerted by the gas.

Have a nice day!

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