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2 years ago

2)

Chemistry

1 answer:

Marina CMI [18]2 years ago

4 0

Answer:

A.moles of solute per liter of solution

Explanation:

the total number of moles of substances present; mass percentage, the ratio of the mass of the solute to the mass of the solution times 100; parts per thousand (ppt), grams.

:-befrank

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If 45.0 mL of a 0.0500 M HNO3, 10.0 mL of a 0.0500 M KSCN, and 30.0 mL of a 0.0500 M Fe(NO3)3 are combined, what is the initial

jeka94

Answer:

the initial concentration of SCN- in the mixture is 0.00588 M

Explanation:

The computation of the initial concentration of the SCN^- in the mixture is as follows:

As we know that

$KSCN \rightarrow K^ + SCN^-$

As it is mentioned in the question that KSCN is present 10 mL of 0.05 M

So, the total milimoles of SCN^- is

= 10 × 0.05

= 0.5 m moles

The total volume in mixture is

= 45 + 10 + 30

= 85 mL

Now the initial concentration of the SCN^- is

= 0.5 ÷ 85

= 0.00588 M

hence, the initial concentration of SCN- in the mixture is 0.00588 M

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3 years ago

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3 years ago

The expression of the theoretical yield (TY) in function of limiting reagent (LR) of a reaction is as follows: TY = ideal mole r

spin [16.1K]

Answer: The theoretical yield of acetanilide is 6.5 grams.

Explanation:

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

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

For aniline:

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:

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

For acetic anhydride:

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.

3 0

3 years ago