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GREYUIT [131]
3 years ago
6

Strontium phosphate reacts with sulfuric acid to form strontium sulfate and phosphoric acid. what is the coefficient for sulfuri

c acid when the equation is balanced using the lowest, whole-numbered coefficients?
Chemistry
2 answers:
Svetllana [295]3 years ago
6 0

<u>Answer:</u> The coefficient of sulfuric acid in the given equation is 3.

<u>Explanation:</u>

The chemical equation for the reaction of strontium phosphate and sulfuric acid follows:

Sr_3(PO_4)_2+3H_2SO_4\rightarrow 3SrSO_4+2H_3PO_4

By Stoichiometry of the reaction:

1 mole of strontium phosphate reacts with 3 moles of sulfuric acid to produce 3 moles of strontium sulfate and 2 moles of phosphoric acid.

The given reaction is a type of double displacement reaction because here exchange of ions takes place.

Hence, the coefficient of sulfuric acid in the given equation is 3.

Veseljchak [2.6K]3 years ago
4 0
The lowest number to this question is 2
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What acid and what base would you choose to prepare the salt potassium perchlorate (kclo4)?
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Can some one help me out? I can't chose between A or B
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3 years ago
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Given the equilibrium constants for the following two reactions at a 298K:NiO(s) + H2(g) ⇌ Ni(s) + H2O(g) Kc=40NiO(s) +CO(g) ⇌ N
shusha [124]

Answer:

The value  is  K_C =  \frac{40}{600}

Explanation:

From the question

   The equation given is  

            NiO(s) + H2(g) ⇌ Ni(s) + H2O(g) Kc=40          (1)

            NiO(s) +CO(g) ⇌ Ni(s) +CO2(g) Kc=600         (2)

Generally the reverse of the second equation as shown below

            Ni(s) +CO2(g) ⇌  NiO(s) + CO(g)                 (3)

The equilibrium constant becomes    K_c  ' =  \frac{1}{600}  

 Now  adding  1  and  3 we obtain

      NiO(s) + H2(g)+ Ni(s) +CO2(g) ⇌ Ni(s) + H2O(g)+NiO(s) + CO(g)        

               CO2(g) + H2(g) ⇌ CO(g) + H2O(g)

Hence the equilibrium constant for the resulting equation is mathematically evaluated as

           K_C =  K_c' *  K_c

=>       K_C =  \frac{1}{600}  *  40

=>         K_C =  \frac{40}{600}

       

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3 years ago
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