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3 years ago

25 POINTS FOR SCIENCE! PLEASE HELP!! THANK YOU!!!!

Chemistry

1 answer:

astra-53 [7]3 years ago

7 0

1. 9.36 moles S

2. 0.213 moles Al

<h3>Further explanation</h3>

Given

Reaction

2Al+3S⇒Al₂S₃

Required

moles of S

moles of Al

Solution

The reaction coefficient in a chemical equation shows the mole ratio of the compounds involved in the reaction (reactants or products)

1. 3.12 moles Al₂S₃

From equation, mol ratio of Al₂S₃ : S = 1 : 3, so mol S :

= 3/1 x mol Al₂S₃

= 3/1 x 3.12

= 9.36 moles S

2. 0.32 moles S

From equation, mol ratio of Al : S = 2 : 3, so mol Al :

= 2/3 x mol S

= 2/3 x 0.32

= 0.213 moles Al

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The question is incomplete, the complete question is:

The solubility of slaked lime, $Ca(OH)_2$ , in water is 0.185 g/100 ml. You will need to calculate the volume of $2.50\times 10^{-3}$ M HCl needed to neutralize 14.5 mL of a saturated

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<u>Explanation:</u>

Given values:

Solubility of $Ca(OH)_2$ = 0.185 g/100 mL

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Using unitary method:

In 100 mL, the mass of $Ca(OH)_2$ present is 0.185 g

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The equation used is:

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

Given mass of $Ca(OH)_2$ = 0.0268 g

Molar mass of $Ca(OH)_2$ = 74 g/mol

Plugging values in equation 1:

$\text{Moles of }Ca(OH)_2=\frac{0.0268g}{74g/mol}=0.000362 mol$

Moles of $OH^-$ present = $(2\times 0.000362)=0.000724mol$

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$Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O$

By the stoichiometry of the reaction:

Moles of $OH^-$ = Moles of $H^+$ = 0.000724 mol

The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}$ .....(2)

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Putting values in equation 2, we get:

$2.50\times 10^{-3}mol=\frac{0.000724\times 1000}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.000725\times 1000}{2.50\times 10^{-3}}=290mL$

Hence, the volume of HCl required is 290mL, the mass of $Ca(OH)_2$ is 0.0268g, the moles of

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