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Kryger [21]
3 years ago
5

Why is the chemical formula Li2H incorrect? Select the correct answer below: A. There should be one lithium, not two. B. There s

hould be one lithium and two hydrogens. C. Lithium does not react with hydrogen. D. There should be three lithiums, not two.
Chemistry
1 answer:
gogolik [260]3 years ago
3 0

Answer:

There should be one lithium, not two.

Explanation:

Lithium reacts with hydrogen at about 750°C to yield lithium hydride (LiH). LiH is white and powdery in appearance. It releases hydrogen gas when it reacts with water.

The correct formula for Lithium hydride is LiH and not Li2H because both lithium and hydrogen are univalent. Lithium has a valency of +1 while hydrogen has a valency of -1 in lithium hydride. Hydrides are formed between hydrogen and highly electro positive metals. In hydrides, hydrogen is forced to accept an electron from the highly electro positive metal.

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A 1.00 kg sample of Sb2S3 (s) and a 10.0 g sample of H2 (g) are allowed to react in a 25.0 L container at 713 K. At equilibrium,
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<u>Answer:</u> The value of K_c is coming out to be 0.412

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For Sb_2S_3</u>

Given mass of Sb_2S_3 = 1.00 kg = 1000 g   (Conversion factor: 1 kg = 1000 g)

Molar mass of Sb_2S_3 = 339.7 g/mol

Putting values in equation 1, we get:

\text{Moles of }Sb_2S_3=\frac{1000g}{339.7g/mol}=2.944mol

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 10.0 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

\text{Moles of hydrogen gas}=\frac{10.0g}{2g/mol}=5mol

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Given mass of hydrogen sulfide = 72.6 g

Molar mass of hydrogen sulfide = 34 g/mol

Putting values in equation 1, we get:

\text{Moles of hydrogen sulfide}=\frac{72.6g}{34g/mol}=2.135mol

The chemical equation for the reaction of antimony sulfide and hydrogen gas follows:

                  Sb_2S_3(s)+3H_2(g)\rightarrow 2Sb(s)+3H_2S(g)

Initial:            2.944      5

At eqllm:      2.944-x     5-3x         2x        3x

We are given:

Equilibrium moles of hydrogen sulfide = 2.135 moles

Calculating for 'x', we get:

\Rightarrow 3x=2.135\\\\\Rightarrow x=\frac{2.135}{3}=0.712

Equilibrium moles of hydrogen gas = (5 - 3x) = (5 - 3(0.712)) = 2.868 moles

Volume of the container = 25.0 L

Molarity of a solution is calculated by using the formula:

\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}

The expression of K_c for above equation, we get:

K_c=\frac{[H_2S]^3}{[H_2]^3}

The concentration of solids and liquids are not taken in the expression of equilibrium constant.

K_c=\frac{(\frac{2.135}{25})^3}{(\frac{2.868}{25})^3}\\\\K_c=0.412

Hence, the value of K_c is coming out to be 0.412

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