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1 year ago

Calculate the enthalpy of the reaction below (∆Hrxn, in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g).

1 answer:

7 0

The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

The bond energies data is given as follows:

BE for C≡O = 1072 kJ/mol

BE for Cl-Cl = 242 kJ/mol

BE for C-Cl = 328 kJ/mol

BE for C=O = 766 kJ/mol

The enthalpy change for the reaction is given as :

ΔHr×n = ∑H reactant bond - ∑H product bond

ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )

ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )

ΔHr×n = 1314 - 1422

ΔHr×n = - 108 kJ

Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

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3 years ago

Read 2 more answers

What is the percent yield of LiCl if I produced 30.85g LiCl and my theoretical yield was calculated to be 35.40g LiCl?

marissa [1.9K]

Answer:

87.15%

Explanation:

To find percent yield, we can use this simple equation

$\frac{Actual}{Theoretical} *100$

Where "Actual" is the amount in grams actually collected from the reaction, and "Theoretical" is, well, the theoretical amount that should have been produced.

They give us these values, so to find the percent yield, just plug the numbers in.

$\frac{30.85}{35.40} *100\\\\ =87.15$

So, the percent yield is 87.15%

An easy trick to remember how to do this is just to divide the smaller number by the bigger number and move the decimal back two places. If you have a percent yield greater than 100%, something is wrong in the reaction.

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3 years ago

What mass of HCL, in grams, is required to react with 0.610 g of al(oh)3 ?

kompoz [17]

Answer: 0.8541 grams of HCl will be required.

Explanation: Moles can be calculated by using the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of $Al(OH)_3$ = 0.610 g

Molar mass of $Al(OH)_3$ = 78 g/mol

$\text{Number of moles}=\frac{0.610g}{78g/mol}$

Number of moles of $Al(OH)_3$ = 0.0078 moles

The reaction between $Al(OH)_3$ and HCl is a type of neutralization reaction because here acid and base are reacting to form an salt and also releases water.

Chemical equation for the above reaction follows:

$Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O$

By Stoichiometry,

1 mole of $Al(OH)_3$ reacts with 3 moles of HCl

So, 0.0078 moles of $Al(OH)_3$ will react with $\frac{3}{1}\times 0.0078$ = 0.0234 moles

Mass of HCl is calculated by using the mole formula, we get

Molar mass of HCl = 36.5 g/mol

Putting values in the equation, we get

$0.0234moles=\frac{\text{Given mass}}{36.5g/mol}$

Mass of HCl required will be = 0.8541 grams

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