How many electrons are needed for cio3-

In the context of this problem, we have a chemical reaction between hydrochloric acid and calcium hydroxide. HCl is the acid here and calcium hydroxide is the base. Hence, we have an acid-base reaction, also known as neutralization reaction.

In a neutralization reaction, water is produced as a product, as well as a salt that we obtain after we exchange the cations: calcium bonds to chloride and hydrogen bonds to hydroxide (the latter is the formation of water). This means we also produce calcium chloride as a product. The overall reaction represents this as:

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

Firslt of all, we wish to find the number of moles of HCl present. Having molarity and volume, this is done by applying the molarity formula. It states that molarity is equal to the rate between moles and volume:

$c_{HCl}=\frac{n_{HCl}}{V_{HCl}}$

Rearranging for moles of HCl, n:

$n_{HCl}=c_{HCl}V_{HCl}$

Based on stoichiometry of the balanced chemical equation, notice that 1 mole of calcium hydroxide reacts with 2 moles of HCl, meaning:

$n_{Ca(OH)_2}=\frac{1}{2} n_{HCl}=\frac{1}{2}c_{HCl}V_{HCl}$

Now that we have the expression for moles, we may also express moles of calcium hydroxide as the ratio between its mass and molar mass:

$n_{Ca(OH)_2}=\frac{m_{Ca(OH)_2}}{M_{Ca(OH)_2}}$

Using the last two equations, we see that:

$\frac{1}{2}c_{HCl}V_{HCl}=\frac{m_{Ca(OH)_2}}{M_{Ca(OH)_2}}\\\therefore m_{Ca(OH)_2}=\frac{1}{2}c_{HCl}V_{HCl}M_{Ca(OH)_2}$

Substitute the given data, as well as the molar mass of calcium hydroxide:

$m_{Ca(OH)_2}=\frac{1}{2}\cdot0.208 M\cdot0.0427 L\cdot74.093 g/mol=0.329 g$

8 0

4 years ago

34g aluminum are combined with 39g chlorine gas which is the limiting reactant

vampirchik [111]

The limiting reactant is chlorine (Cl2).

<u>Explanation</u>:

Limiting reactant is the amount of product formed which gets limited by the reagent without continuing it.

2 Al + 3 Cl2 ==> 2 AlCl3 represents the balanced equation.

Number of moles Al present = 34 g Al x 1 mole Al / 26.98 g

= 1.260 g moles of Al

Number of moles Cl2 present = 39 g Cl2 x 1 mole Cl2 / 35.45 g

= 1.10 g moles of Cl2

Dividing each reactant by it's coefficient in the balanced equation obtains:

1.260 moles Al / 2 = 0.63 g moles of Al

1.11 moles Cl2 / 3 = 0.36 g moles of Cl2

The reactant which produces a lesser amount of product is called as limiting reactant.

Here the Limiting reactant is Cl2.

6 0

3 years ago

The buffer solution is used to control the pH to insure that it does not become too high because excessively basic solutions cou

oksian1 [2.3K]

Answer:

pH = 12.7

Explanation:

First, we have to calculate the [Ca²⁺] in a solution of about 250 ppm CaCO₃.

$\frac{250mgCaCO_{3}}{L} .\frac{1gCaCO_{3}}{1000mgCaCO_{3}} .\frac{1molCa^{2+} }{100gCaCO_{3}} =2.5 \times 10^{-3} M$

Now, let's consider the dissolution of Ca(OH)₂ in water.

Ca(OH)₂(s) ⇄ Ca²⁺(aq) + 2 OH⁻(aq)

The solubility product Ksp is:

Ksp = [Ca²⁺] × [OH⁻]²

[OH⁻] = √(Ksp/[Ca²⁺]) = √(6.5 × 10⁻⁶/2.5 × 10⁻³) = 5.1 × 10⁻² M

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (5.1 × 10⁻²) = 1.3

pH + pOH = 14 ⇒ pH = 14 - pOH = 14 - 1.3 = 12.7

5 0

3 years ago