<u>Answer:</u> The chemical equation is 
<u>Explanation:</u>
There are 2 types of reactions that are classified based on enthalpy change:
- Endothermic reaction
- Exothermic reaction
Endothermic reactions: They are defined as the reactions where heat is absorbed by the reaction. The change in enthalpy of the reaction is always positive.
Exothermic reactions: They are defined as the reactions where heat is released by the reaction. The change in enthalpy of the reaction is always negative.
Given values:
Energy released for 1 mole of HCl reacted = -42.1 kCal
The chemical equation for the formation of ammonium chloride follows:
The percent by mass of nitric acid in the mixture is 3.36 %
2 HNO3 + Ba(OH)2 --->Ba(NO3)2 + 2H2O
so 1 mole Ba(OH)2 neutralizes 2 mol HNO3
moles of Ba(OH)2 present = molarity of base* volume of base
= 0.229 M * 15.6 ml
= 3.5724 milli mols
mols of HNO3 neutralized = 2* mols of Ca(OH)2 used
= 2* 3.5724 milli mols = 7.1448 x10^-3 mols
molar mass of HNO3 = 63.01 g/mol
mass of HNO3 present = molar mass * mols of HNO3
= 63.01 g/mol * 7.1448 x10^-3 mols
=0.4502 g
mass of sample = 13.4 g
mass % = mass of HNO3/mass of sample * 100
= 0.4502/13.4 *100
= 3.36 %
Nitric acid is a colorless, fuming, and distinctly corrosive liquid that could be a not unusual laboratory reagent and an important commercial chemical for the manufacture of fertilizers and explosives.
Learn more about Nitric acid here: brainly.com/question/22698468
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Answer is: 9.69 grams of ethyl butyrate would be synthesized.
Chemical reaction: C₃H₇COOH + C₂H₅OH → C₃H₇COOC₂H₅ + H₂O.
m(C₃H₇COOH) = 7.35 g; mass of butanoic acid.
n(C₃H₇COOH) = m(C₃H₇COOH) ÷ M(C₃H₇COOH).
n(C₃H₇COOH) = 7.35 g ÷ 88.11 g/mol.
n(C₃H₇COOH) = 0.0834 mol; amount of butanoic acid.
From chemical reaction: n(C₃H₇COOH) : n(C₃H₇COOC₂H₅) = 1 : 1.
n(C₃H₇COOC₂H₅) = 0.0834 mol; amount of ethyl butyrate.
m(C₃H₇COOC₂H₅) = n(C₃H₇COOC₂H₅) · M(C₃H₇COOC₂H₅).
M(C₃H₇COOC₂H₅) = 6·Ar(C) + 2·Ar(O) + 12·Ar(H) · g/mol.
M(C₃H₇COOC₂H₅) = 6·12.01 + 2·15.99 + 12·1.01 · g/mol.
M(C₃H₇COOC₂H₅) = 116.16 g/mol; molar mass of ethyl butyrate.
m(C₃H₇COOC₂H₅) = 0.0834 mol · 116.16 g/mol.
m(C₃H₇COOC₂H₅) = 9.69 g.
