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azamat
3 years ago
15

Which group is more reactive than the alkaline earth metals?

Chemistry
2 answers:
SIZIF [17.4K]3 years ago
8 0

Answer: Alkaline metals(Group 1)

Explanation:

These elements react vigorously with cold water to give hydrogen gas and they have one electron in their outer-most shell which allows them to give electron freely and are good reducing agents.

examples of these elements are:

Hydrogen

Lithium

Sodium

Potassium

Rubidium

Hydrogen is the only gas in group 1.

Vlad [161]3 years ago
5 0

Answer: <em>The </em><em>alkali metals</em><em> reactivity is generally higher than the alkaline earth metals. </em>

Explanation: <em>This makes the alkaline earth metals with their two valence electrons less reactive than alkali metals with their one valence electron.</em>

<u>Alkali metals is ns1 and alkaline earth metals is ns2</u>

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The expression of the theoretical yield (TY) in function of limiting reagent (LR) of a reaction is as follows: TY = ideal mole r
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<u>Answer:</u> The theoretical yield of acetanilide is 6.5 grams.

<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 aniline:</u>

Given mass of aniline = 4.50\times 10^0=4.50g      (We know that:  10^0=1 )

Molar mass of aniline = 93.13 g/mol

Putting values in equation 1, we get:

\text{Moles of aniline}=\frac{4.50g}{93.13g/mol}=0.048mol

  • <u>For acetic anhydride:</u>

To calculate the mass of acetic anhydride, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Volume of acetic anhydride = (1.25\times \text{Mass of aniline})=1.25\times 4.50=5.625mL

Density of acetic anhydride = 1.08 g/mL

Putting values in above equation:

1.08g/mL=\frac{\text{Mass of acetic anhydride}}{5.625mL}\\\\\text{Mass of acetic anhydride}=(1.08g/mL\times 5.625mL)=6.08g

Given mass of acetic anhydride = 6.08 g

Molar mass of acetic anhydride = 102.1 g/mol

Putting values in equation 1, we get:

\text{Moles of acetic anhydride}=\frac{6.08g}{102.1g/mol}=0.06mol

The chemical equation for the reaction of aniline and acetic anhydride follows:

C_6H_5NH_2+CH_3COOCOCH_3\rightarrow C_6H_5NHCOCH_3+CH_3COOH

By Stoichiometry of the reaction:

1 mole of aniline reacts with 1 mole of acetic anhydride

So, 0.048 moles of aniline will react with = \frac{1}{1}\times 0.048=0.048mol of acetic anhydride

As, given amount of acetic anhydride is more than the required amount. So, it is considered as an excess reagent.

Thus, aniline is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of aniline produces 1 mole of acetanilide

So, 0.048 moles of aniline will produce = \frac{1}{1}\times 0.048=0.048mol of acetanilide

Now, calculating the theoretical yield of acetanilide by using equation 1:

Moles of acetanilide = 0.048 moles

Molar mass of acetanilide = 135.17 g/mol

Putting values in equation 1, we get:

0.048mol=\frac{\text{Mass of acetanilide}}{135.17g/mol}\\\\\text{Mass of acetanilide}=(0.048mol\times 135.17g/mol)=6.5g

Hence, the theoretical yield of acetanilide is 6.5 grams.

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