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3 years ago

Limiting Reagent

Chemistry

1 answer:

Bumek [7]3 years ago

8 0

Answer:

a) the percentage yield will exceed 100%

b) the excess reactant is filtered along with the barium sulphate precipitate. It is possible to recover the excess reactant by carefully washing the precipitate with water.

Explanation:

In the precipitation of barium sulphate, the ions in the reactants exchange partners in the product leading to an insoluble product.

In every reaction, there is a limiting reactant whose amount determines the amount of product that can be obtained. The reactant in excess remains in the system even after the reaction is completed and may be recovered alongside the product which leads to a percentage yield above 100%.

If the excess reactant is soluble in water, it can be recovered from the precipitate if needed by washing the precipitate with water.

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Consider the following weak acids and their Ka values: Acetic Acid Ka = 1.8×10^−5 Phosphoric Acid Ka = 7.5×10^−3 Hypochlorous Ac

vfiekz [6]

Answer:

a. Phosphoric Acid

b. Acetic Acid

c. Hypochlorous Acid

Explanation:

A buffer works when the pH of this one is in pKa ± 1. That means, to find which buffer system works in some pH you need to find pKa:

pKa = -log Ka

pKa Acetic acid:

-log1.8x10⁻⁵ = 4.74

pKa phosphoric acid:

-log7.5x10⁻³ = 2.12

pKa hypochlorous acid:

-log3.5x10⁻⁸ = 7.46

a. For a pH of 2.8 the best choice is phophoric acid because its effective range is: 1.12 - 3.12 and 2.8 is between these values.

b. pH 4.5. Acetic acid. effective between pH's 3.74 - 5.74

c. pH 7.5. Hypochlorous acid that works between 6.46 and 8.46

7 0

3 years ago

Which sub-atomic particle is NOT found at the center of the atom?

Ostrovityanka [42]

All the positive charge of an atom is contained in the nucleus, and originates from the protons. Neutrons are neutrally-charged. Electrons, which are negatively-charged, are located outside of the nucleus.

8 0

4 years ago

Read 2 more answers

4. Chemical formula, mg (Cl0₃)2

grin007 [14]

Answer:

0.017mole

0.0033M

Explanation:

Given parameters:

Formula of the compound:

Mg(ClO₃)₂

Mass of the sample = 3.24g

Unknown:

Number of moles of the sample = ?

Molarity = ?

Solution:

The number of moles of any substance is given as:

Number of moles = $\frac{mass}{volume}$

Molar mass of Mg(ClO₃)₂ = 24 + 2[35.5 + 3(16)] = 191g/mol

Number of moles = $\frac{3.24}{191}$ = 0.017mole

Molarity is the number of moles of a solute in a solution:

Molarity = $\frac{number of moles }{volume}$

Volume given = 5.08L

Molarity = $\frac{0.017}{5.08}$ = 0.0033M

5 0

3 years ago

For the chemical reaction CaI2+2AgNO3-> 2AgI+Ca(NO3)2 how many moles of silver iodide will be produced from 205g of calcium i

Colt1911 [192]

Answer:

Moles of silver iodide produced = 1.4 mol

Explanation:

Given data:

Mass of calcium iodide = 205 g

Moles of silver iodide produced = ?

Solution:

Chemical equation:

CaI₂ + 2AgNO₃ → 2AgI + Ca(NO₃)₂

Number of moles calcium iodide:

Number of moles = mass/ molar mass

Number of moles = 205 g/ 293.887 g/mol

Number of moles = 0.7 mol

Now we will compare the moles of calcium iodide with silver iodide.

CaI₂ : AgI

1 : 2

0.7 : 2×0.7 = 1.4

Thus 1.4 moles of silver iodide will be formed from 205 g of calcium iodide.

6 0

3 years ago

How many grams of ammonia (NH3) can be produced by the synthesis of excess hydrogen gas (H2) and 253.8 grams of nitrogen gas (N2

kogti [31]

Answer:

308.2 g of NH₃.

Explanation:

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

3H₂ + N₂ —> 2NH₃

Next, we shall determine the mass of N₂ that reacted and the mass of NH₃ produced from the balanced equation. This can be obtained as follow:

Molar mass of N₂ = 2 × 14 = 28 g/mol

Mass of N₂ from the balanced equation = 1 × 28 = 28 g

Molar mass of NH₃ = 14 + (3×1)

= 14 + 3 = 17 g/mol

Mass of NH₃ from the balanced equation = 2 × 17 = 34 g

Summary:

From the balanced equation above,

28 g of N₂ reacted to produce 34 g of NH₃.

Finally, we shall determine the mass of NH₃ produced by the reaction of 253.8 g of N₂. This can be obtained as illustrated below:

From the balanced equation above,

28 g of N₂ reacted to produce 34 g of NH₃.

Therefore, 253.8 g of N₂ will react to produce = (253.8 × 34)/28 = 308.2 g of NH₃.

Thus, 308.2 g of NH₃ were obtained from the reaction.

8 0

3 years ago