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Gennadij [26K]
3 years ago
8

Of the five salts listed below, which has the highest concentration of its cation in water? assume that all salt solutions are s

aturated and that the ions do not undergo any additional reactions in water.

Chemistry
1 answer:
Zigmanuir [339]3 years ago
5 0
First of all, I need to know what these five salts are. Luckily, I found a similar problem from another website which is shown in the attached picture. The Ksp is the solubility product constant. It follows the formula:

Ksp = [cation concentration]ᵃ[anion concentration]ᵇ
where a and b are the subscripts of the metal and nonmetal, respectively. 

For the solutions ahead, let x be the concentration of the cation.

A.  The formula is PbCr₂O₄.
2.8×10⁻¹³ = [x][x]
Solving for x, x = 5.29×10⁻⁷ M

B. The formula is Co(OH)₂. 
 1.3×10⁻¹⁵ = [x][x]²
Solving for x, x = 1.09×10⁻⁵ M

C. The formula is CoS. 
 5×10⁻²² = [x][x]
Solving for x, x = 2.24×10⁻¹¹ M

D. The formula is Cr(OH)₃. 
 1.6×10⁻³⁰ = [x][x]³
Solving for x, x = 3.56×10⁻⁶ M

E. The formula is Ag₂S. 
 6×10⁻⁵¹ = [x]²[x]
Solving for x, x = 1.82×10⁻¹⁷ M

<em>Thus,the highest concentration is letter B, Cobalt (II) Hydroxide.</em>

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Nitrogen and hydrogen gases react to form ammonia gas via the following reaction:
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1mol N2 + 3mol H2 → 2mol NH3 
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Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N
Montano1993 [528]

The question is incomplete, here is the complete question:

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.

Determine the limiting reactant. (express your answer as a chemical formula)

<u>Answer:</u> The limiting reactant is ammonia (NH_3)

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

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

  • <u>For ammonia:</u>

Given mass of ammonia = 135.9 kg = 135900 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonia}=\frac{135900g}{17g/mol}=7994.12mol

  • <u>For carbon dioxide gas:</u>

Given mass of carbon dioxide gas = 211.4 kg = 211400 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon dioxide gas}=\frac{211400g}{44g/mol}=4804.54mol

The given chemical reaction follows:

2NH_3(aq.)+CO_2(aq,)\rightarrow CH_4N_2O(aq.)+H_2O(l)

By Stoichiometry of the reaction:

2 moles of ammonia reacts with 1 mole of carbon dioxide

So, 7994.12 moles of ammonia will react with = \frac{1}{2}\times 7994.12=3997.06mol of carbon dioxide

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

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

Hence, the limiting reactant is ammonia (NH_3)

5 0
3 years ago
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