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ratelena [41]
3 years ago
5

Please Help , Thank You

Chemistry
1 answer:
ivanzaharov [21]3 years ago
8 0
I think it enters a new  material
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Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:
Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: Al_{2}S_{3} +H_{2}O  _\to  Al(OH)_{3} + H_{2}S

First, balance the chemical equation

Al_{2}S_{3} + 6H_{2}O  \to 2Al(OH)_{3} + 3H_{2}S

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

n_{Al_{2}S_{3}  } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3}  } = \frac{158g}{150.16g/mol}   \\n_{Al_{2}S_{3} }=1.05 mol

From the chemical reaction , the ratio of molar is 3mol H_{2}S/1 mol Al_{2}S_{3}. So, the moles of hydrogen sulfide are:

n_{H_{2} O} =\frac{131g}{18.02g/mol}

       = 7.26mol

From the chemical reaction, the molar ratio is 3 mol H_{2}S/6 mol H_{2}O. So, the moles of hydrogen sulfide are:

Moles of H_{2}S formed = 7.26 mol H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }

Th liming reactant isAl_{2}S_{3} beacuse the mass of Al_{2}S_{3} forms less product than water. Therefore, the maximum number of moles of H_{2}S is 3.15 mol.  We know that molar mass of H_{2}S is 34.10g/mol. So, the maximum mass of H_{2}S formed is,

m_{H_{2}S } = n_{H_{2}S } \times Molar mass of H_{2}S

         = 3.15 mol \times 34.10g/mol

         = 107.4g

Now, multiplying the number of moles of Al_{2}S_{3} by the molar ratio between Al_2S_3 and H_2O which is 6mol H_2O/1mol Al_2S_3 we get the number of moles of H_2O reacted.

Moles of H_2O reacted = 1.05 mol Al_{2}S_3 \times \frac{6 mol H_2O}{1 mol Al_2S_3}

                                     = 6.31 mol H_2O

The mass of H_2O is,

m_{H_{2} O} = 6.31 mol \times 18.02g/ mol

          = 114g

On subtracting, the mass of H_2O reacted from the given mass of H_2O is,

m_{H_2O} = (131-114)g

         = 17g

Hence, the excess remaining reactant is 17g

To learn more about Chemical Reaction

brainly.com/question/11231920

#SPJ4

 

8 0
2 years ago
What is 30% NaOH test​
pogonyaev

Sodium hydroxide solution 30% CAS 1310-73- 2 Suprapur® - Find MSDS or SDS, a COA, data sheets and more information.

CAS number:

4 0
3 years ago
Instead of 6 M NaOH being added to the solution, 6 M HCl is added. How will this affect the test for the presence of ammonium io
Lera25 [3.4K]

Answer:

Ammonia gas(an alkaline gas with characteristics of choking or irritating smell) is not liberated when 6mole of HCl is added to the solution instead of 6mole of NaOH, to test for the presence of ammonium ion in the solution

Explanation:

As expected, when testing for ammonium ion in a solution (precisely ammonium salt solution), Sodium Hydroxide (NaOH) is required as the test reagent.

When NaOH is added to the solution, A gas with characteristics of choking or irritating smell is liberated.

This gas turn red litmus paper blue.

This liberated gas is an alkaline gas, which is confirmed as an ammonia gas(NH3).

If HCl is added instead of NaOH, the ammonia gas will not be liberated, which indicates that the test reagent used is wrong.

3 0
3 years ago
Without doing any calculations, match the following thermodynamic properties with their appropriate numerical sign for the follo
siniylev [52]

Answer:

∆H > 0

∆Srxn <0

∆G >0

∆Suniverse <0

Explanation:

We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.

Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.

The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.

Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.

3 0
3 years ago
Think about a carbon atom that is released into the atmosphere from the burning of wood in a campfire. If it were to go through
viva [34]
1) <span>Step 5 A tree absorbs the carbon from the atmosphere into its leaves for photosynthesis. Inorganic carbon is turned into organic.
2) </span><span>Step 1 A caterpillar gets the carbon by eating the tree's leaves. Caterpillar use carbon for energy.
3) </span><span>Step 3 A bird gets the carbon by eating the caterpillar. Organic carbon shifts from one animal to another
4) </span><span>Step 4 The bird flies into a building and dies instantly. It falls to the ground.
5) </span><span>Step 2 The bird decomposes and the carbon is added to the atmosphere. Organic carbon turns into inorganic.
</span>
3 0
3 years ago
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