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3 years ago

Given the following reaction:

Chemistry

1 answer:

Wewaii [24]3 years ago

3 0

Answer:

Approx. 4⋅g.

Explanation:

Moles of sulfuric acid =10.0⋅g98.08⋅g⋅mol−1=0.102⋅mol.

Now we have the molar quantity of sulfuric acid that react; we also have the stoichiometric equation that shows the molar equivalence of sulfuric acid, and lithium hydroxide.

Given the stoichiometry,

mass of water =0.102⋅mol×2×18.01.g.mol−1=??⋅g.

Why did I multiply the mass in this equation by 2? Am I pulling your leg?

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How many moles of chloride ions are there in 2 moles of calcium chloride?

Oksana_A [137]

Answer:

There are 4 moles of chloride ions in 2 moles of calcium chloride.

Explanation:

Start by writing the chemical formula of calcium chloride.

The suffix -ide in chloride means that chlorine is working as the ion Cl⁻, this is with oxidation state -1.

Calcium, being an alkaline earth metal, uses oxidation state +2, Ca²⁺

Then, two atoms of Cl are combined with one atom of calcium, because in this way each Cl atom will accept one electron from a Ca atom, which will donate its two valence electrons.

That is indicated in the chemical formula with subscripts: CaCl₂, meaning that each formula-unit of calcium chloride has 2 ions of chlorine and 1 ions of calcium, or twice the number of moles of chloride ion.

1 mol CaCl₂ : 1 mol Ca⁺ : 2 mol Cl⁻

⇒ 2 mol CaCl₂ : 2 mol Ca⁺ : 4 mol Cl⁻

Hence, there are 4 moles of chloride ions in 2 moles of calcium chloride.

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4 years ago

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3 years ago

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A sample of 9.27 g9.27 g of solid calcium hydroxide is added to 38.5 mL38.5 mL of 0.500 M0.500 M aqueous hydrochloric acid. Writ

navik [9.2K]

Answer: The excess reagent for the given chemical reaction is calcium hydroxide and the amount left after the completion of reaction is 0.115375 moles. The amount of calcium chloride formed in the reaction is 1.068 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 calcium hydroxide:

Given mass of calcium hydroxide = 9.27 g

Molar mass of calcium hydroxide = 74.093 g/mol

Putting values in above equation, we get:

$\text{Moles of calcium hydroxide}=\frac{9.27g}{74.093g/mol}=0.125mol$

To calculate the moles of a solute, we use the equation:

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

We are given:

Volume of hydrochloric acid = 38.5mL = 0.0385 L (Conversion factor: 1 L = 1000 mL)

Molarity of the solution = 0.500 moles/ L

Putting values in above equation, we get:

$0.500mol/L=\frac{\text{Moles of hydrochloric acid}}{0.0385L}\\\\\text{Moles of hydrochloric acid}=0.01925mol$

For the given chemical equation:

$2HCl(aq.)+Ca(OH)_2(s)\rightarrow CaCl_2(s)+2H_2O(l)$

Here, the solid salt is calcium chloride.

By Stoichiometry of the reaction:

2 moles of hydrochloric acid reacts with 1 mole of calcium hydroxide.

So, 0.01925 moles of hydrochloric acid will react with = $\frac{1}{2}\times 0.01925=0.009625moles$ of calcium hydroxide.

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

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

Amount of excess reagent (calcium hydroxide) left = 0.125 - 0.01925 = 0.115375 moles

By Stoichiometry of the reaction:

2 moles of hydrochloric acid produces 1 mole of calcium chloride.

So, 0.01925 moles of hydrochloric acid will produce = $\frac{1}{2}\times 0.01925=0.009625moles$ of calcium chloride.

Now, calculating the mass of calcium chloride from equation 1, we get:

Molar mass of calcium chloride = 110.98 g/mol

Moles of calcium chloride = 0.009625 moles

Putting values in equation 1, we get:

$0.009625mol=\frac{\text{Mass of calcium chloride}}{110.98g/mol}\\\\\text{Mass of calcium chloride}=1.068g$

Hence, the excess reagent for the given chemical reaction is calcium hydroxide and the amount left after the completion of reaction is 0.115375 moles. The amount of calcium chloride formed in the reaction is 1.068 grams.

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Calculate the volume (in mL) of a 2.75 M solution that must be used to make 1.25L of a 0.150 M solution.

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