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3 years ago

Stoichiometry

1 answer:

8 0

Stoichiometry is the ratio of reactants to products in a balanced chemical reaction equation.
the stoichiometry in this case of H₂ to O₂ is 2:1.
This means that 2 moles of H₂ reacts with 1 mol of O₂.
we have been asked to calculate the volume of H₂ gas.
Since both O₂ and H₂ are gases, at standard conditions its stated that molar volume of gases is 22.4 L
This means that at standard temperature and pressure, 1 mol of any gas occupies a volume of 22.4 L
first we need to calculate the number of O₂ moles reacted;
22.4 L of gas - 1 mol of O₂
1 L of gas - 1/22.4 mol/L
then 20 L of O₂ - 1/22.4 mol/L * 20 L = 0.89 mol
stoichiometry of H₂ to O₂ is 2:1
the number of H₂ moles = 0.89*2 = 1.79 mol
1 mol occupies 22.4 L
Therefore 1.79 mol = 22.4 L/mol * 1.79 mol = 40 L
Therefore it can be seen that stoichiometry applies to volumes as well.
volume of H₂ : O₂ = 40 L : 20 L = 2:1
volume and moles both can be determined by stoichiometry.
Volume of H₂ reacted = 40 L

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Answer:

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The protons of methylene group between the two carbonyl groups in ethylacetoacetate are acidic in nature. When compounds containing such acidic protons are treated with bases the loose proton and form enolates.

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3 years ago

You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

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Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

Now put all the given values in the above formula, we get:

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

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3 years ago

Select the correct hybridization for the central atom based on the electron geometry cocl2 (carbon is the central atom).

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In this compound (Phosgene) the central atom (carbon is Sp² Hybridized).

Sp, Sp² and Sp³ can be calculated very simply by doing three steps,

Step 1:
Assume triple bond and double bond as one bond and assign s or p to it. In this example carbon double bond oxygen is considered once and let suppose it is s. Now we are having our s.

Step 2:
Count lone pair of electron, each lone pair counts for s and p. In this case there is no lone pair of electron on carbon, so not included.

Step 3:
Count single bonds for s and p. As we have already assigned s to the double bond, now one p for one single bond, and other p for the other single bond.

Result:
So, we counted 1 s for double bond, 1 p for one single and other p for second single bond. As a whole we got,

Sp²

Practice:
You can practice for hybridization of Oxygen in this molecule. Oxygen has 2 lone pair of electrons. (Hint: Sp² Hybridization)

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Answer is: lithium is the least reactive and potassium is the most reactive of the three alkali elements.
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3 years ago