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Greeley [361]
2 years ago
12

How many grams of NaF form when .5 mol of HF reacts with excess Na2SiO3?

Chemistry
1 answer:
IgorLugansk [536]2 years ago
7 0

Answer:

5.25g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is shown below:

Na2SiO3 + 8HF → H2SiF6 + 2NaF + 3H2O

From the balanced equation above,

8 moles of HF reacted to produce 2 moles of NaF.

Therefore, 0.5 moles of HF will react to produce = (0.5 x 2)/8 = 0.125 mole of NaF.

Next, we shall convert 0.125 mole of NaF to grams.

This is illustrated below:

Mole of NaF = 0.125 mole

Molar mass of NaF = 23 + 19 = 42g/mol

Mass of NaF =..?

Mass = mole x molar mass

Mass of NaF = 0.125 x 42

Mass of NaF = 5.25g

Therefore, 5.25g of NaF is produced from the reaction.

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2 years ago
Hydrofluoric acid and Water react to form fluoride anion and hydronium cation, like this HF(aq) + H_2O(l) rightarrow F(aq) + H_3
maksim [4K]

Answer:

Kc = 1.09x10⁻⁴

Explanation:

<em>HF = 1.62g</em>

<em>H₂O = 516g</em>

<em>F⁻ = 0.163g</em>

<em>H₃O⁺ = 0.110g</em>

<em />

To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:

Kc = [H₃O⁺] [F⁻] / [HF]

<em>Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:</em>

<em />

[H₃O⁺] = 0.110g * (1mol /19.01g) = 0.00579moles / 5.6L = 1.03x10⁻³M

[F⁻] = 0.163g * (1mol /19.0g) = 0.00858moles / 5.6L = 1.53x10⁻³M

[HF] = 1.62g * (1mol /20g) = 0.081moles / 5.6L = 0.0145M

Kc = [1.03x10⁻³M] [1.53x10⁻³M] / [0.0145M]

<h3>Kc = 1.09x10⁻⁴</h3>
7 0
3 years ago
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