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svet-max [94.6K]
3 years ago
5

A compound is formed when 9.03 g

Chemistry
1 answer:
NikAS [45]3 years ago
3 0

The percent composition  of this compound :

Mg = 72.182%

N = 27.818%

<h3>Further explanation</h3>

Given

9.03 g  Mg

3.48 g  N

Required

The percent composition

Solution

Proust stated the Comparative Law that compounds are formed from elements with the same Mass Comparison so that the compound has a fixed composition of elements

Total mass of the compound :

= 9.03 g + 3.48 g

= 12.51 g

The percent composition :

Mg : 9.03/ 12.51 g x 100% = 72.182%

N : 3.48 / 12.51 g x 100% = 27.818%

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<u>Answer:</u> The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

<u>For NaOH:</u>

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Putting values in equation 1, we get:

0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol

<u>For sulfuric acid:</u>

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol

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2 moles of NaOH reacts with 1 mole of sulfuric acid

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

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

Excess moles of sulfuric acid = (4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol

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2 moles of KOH produces 1 mole of sodium sulfate

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Moles of sodium sulfate = 1.462\times 10^{-3}moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

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Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

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Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

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