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Airida [17]
3 years ago
14

In an exothermic reaction, chlorine reacts with 2.0200g of hydrogen to form 72.926g of hydrogen chloride gas. how many grams of

chlorine reacted with hydrogen
Chemistry
1 answer:
Artemon [7]3 years ago
4 0

Answer : 70.906 grams of chlorine reacted with hydrogen

Explanation :

Step 1 : Write balanced chemical equation.

The balanced chemical equation for the reaction between hydrogen and chlorine gas is given below.

H_{2} (g) + Cl_{2} (g) \rightarrow 2HCl(g)

Step 2 : Find moles of H₂ gas.

The moles of H₂ can be found as

mole = \frac{grams}{MolarMass}

We have 2.0200 g of H₂ and molar mass of H₂ is 2.02 g/mol.

Let us plug in these values to find moles of H₂.

mole =\frac{2.0200g}{2.02g/mol} = 1 mol

We have 1 mol of H₂.

Step 3 : Find moles of Cl₂ using mole ratio.

The mole ratio of H₂ and Cl₂ is 1 : 1.

The moles of Cl₂ can be calculated as

1 mol H_{2} \times\frac{1 mol Cl_{2}}{1 mol H_{2}} = 1 mol Cl_{2}

Step 4 : Find grams of Cl₂.

Molar mass of Cl₂ gas is 70.096 g/mol

Mass of Cl₂ = 1 mol Cl_{2} \times\frac{70.906g}{mol} = 70.906 g

We have 70.906 grams of Cl₂

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Answer:

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Molecular formula of Isoflurane is <span><span>C</span></span>₃H₂ClF₅O.

Now calculate the percent composition by mass, which means percent of each element in the compound.

Mass of Isoflurane = 184.49 g/mol

Mass of carbon in the compound = 3 x 12.011 = 36.033g

Mass of hydrogen in the compound = 2 x 1.008 = 2.016g

Mass of chlorine in the compound = 1 x 35.453 = 35.453 g

Mass of fluorine in the compound = 5 x 18.998 = 94.99g

Mass of Oxygen in the compound = 1 x 16 = 16 g

Carbon’s percentage = Mass of carbon in the compound /mass of isoflurane x 100 =36.033/184.49 x 100 =19.53%

Hydrogen’s Percentage = Mass of hydrogen in the compound/mass of isoflurane x 100 = 2.016/184.49 = 1.09%

Chlorine’s percentage = Mass of chlorine in the compound/mass of isoflurane x 100 = 35.453/184.49 =19.22%

Flourine’s percentage = Mass of fluorine in the compound/mass of isoflurane x 100 = 94.99/184.49 x 100 = 51.49%

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The first excited state of Ca is reached by absorption of 422.7 nm light. Find the energy difference (kJ/mole) between the groun
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For the excited state of Ca at the absorption of 422.7 nm light,the energy difference  is mathematically given as

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<h3>What is the energy difference (kJ/mole) between the ground and the first excited state?</h3>

Generally, the equation for the Energy  is mathematically given as

E = nhc / λ

Where

h= plank's constant

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c = speed of light

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Therefore

E = 1*(6.625x 10-34 Js)( 3x 10^8 m/s) / ( 422.7x10^-9)

E= 4.70x10-22 kJ/mol

In conclusion, Energy  

E= 4.70x10-22 kJ/mol

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A sample of oxygen gas in one container has a volume of 20.0mililiter at 297 K and 101.3 kPa. The entire sample is transferred t
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Answer: V_2=\frac{101.3kPa\times 20.0ml\times 283K}{297K\times 94.6kPa}

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The combined gas equation is,:

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Now put all the given values in the above equation, we get the final volume of gas.

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