Balanced Chemical Equation Reaction Type: At completion of reactions: Grams of calcium hydroxide Grams of sulfuric acid Grams of water Grams of calcium sulfate
Ionic Equilibrium
Chemical equilibrium and ionic equilibrium are two major concepts in chemistry. Ionic equilibrium deals with the equilibrium involved in an ionization process while chemical equilibrium deals with the equilibrium during a chemical change. Ionic equilibrium is established between the ions and unionized species in a system. Understanding the concept of ionic equilibrium is very important to answer the questions related to certain chemical reactions in chemistry.
Arrhenius Acid
Arrhenius acid act as a good electrolyte as it dissociates to its respective ions in the aqueous solutions. Keeping it similar to the general acid properties, Arrhenius acid also neutralizes bases and turns litmus paper into red.
Bronsted Lowry Base In Inorganic Chemistry
Bronsted-Lowry base in inorganic chemistry is any chemical substance that can accept a proton from the other chemical substance it is reacting with.
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![Balanced Chemical Equation
Reaction Type:
At completion of reactions:
Grams of calcium hydroxide
Grams of sulfuric acid
Grams of water
Grams of calcium sulfate](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F21b31cbf-edf5-4459-b2b9-aeaac7c813ec%2F120ffedb-dcbf-413d-9616-dacf31980ea4%2Fwk7ienr_processed.jpeg&w=3840&q=75)
![Ca(OH)2 (aq) + H₂SO4 (aq) → + 2 H₂O (1) + CaSO4 (aq) (This is LIMITING REACTANT: H₂SO4 is the Limiting Reactant)
Reaction Type: Neutralization
Grams of Product (List the Product and the Amount): 0.180 g H₂O and 0.681 g CaSO4
? g H₂O = 1.839 g Ca(OH)2 x 1 mol Ca(OH)2 x 2 mol H₂O x 18.02 g H₂O = 0.8944 g H₂O
74.10 g Ca(OH)2 1 mol Ca(OH)2
1 mol H₂O
? g H₂O = 25.0 mL H₂SO4 x
1L H₂SO4 x 0.200 mol H₂SO4 x 2 mol H₂O x 18.02 g H₂O = 0.180 g H₂0
1000 mL H₂SO4 IL H₂SO4 1 mol H₂SO4 1 mol H₂O
ONLY USE THE Limiting Reactant TO DETERMINE THE AMOUNT OF THE REMAINING PRODUCT(S).
? g H₂O = 25.0 mL H₂SO4 x
IL H₂SO4 x 0.200 mol H₂SO4 x 1 mol CaSO4 x 136.14 g CaSO4
1000 mL H₂SO4 1L H₂SO4 1 mol H₂SO4 1 mol CaSO4
? g Ca (OH)₂ USED
= 25.0 mLH₂ SO4 X-
= 0.3705 g Ca (OH)2
Amount of Ca(OH)2 remaining in the container
——
IL H₂ SOA
1000 mL H₂ SO4
= Ca (OH)₂ amount given
= 1.839 g GIVEN - 0.3705 g USED
= 1.469 g
0.200 mol H₂ SO4
X-
1 LH₂ SOA
-
X
LEFT OVER = EXCESS
1 mol Ca (OH)₂
1 mol H₂ SO4
Ca (OH)2 amount used
X
0.681 g CaSO4
74.093 g 1 mol Ca (OH)₂2
1 mol Ca (OH)₂](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F21b31cbf-edf5-4459-b2b9-aeaac7c813ec%2F120ffedb-dcbf-413d-9616-dacf31980ea4%2Fmq7wrr_processed.jpeg&w=3840&q=75)
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