The enthalpy of this reaction is -5315 KJ/mol.
The equation of the reaction is;
2C4H10(g) + 13O2(g) -----> 8CO2 (g) + 10H2O(g)
We know that the enthalpy of reaction can be obtained from the enthalpy of formation of the reactants and products as follows;
ΔHrxn = ΔHf(products) - ΔHf(reactants)
We have the following information from the question;
ΔHf C4H10 = - 125. 6 kJ/mol
ΔHf CO2 = - 393. 5 kJ/mol
ΔHf H2O = - 241. 82 kJ/mol
ΔHf O2 = 0 KJ/mol
Hence;
[(8 × (- 393. 5 )) + (10 × (- 241. 82))] - [2( - 125. 6))]
= -5315 KJ/mol
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Answer:
The answer to your question is 41.6 g of AgCl
Explanation:
Data
mass of NH₄Cl = 15.5 g
mass of AgNO₃ = excess
mass of AgCl = 35.5 g
theoretical yield = ?
Process
1.- Write the balanced chemical reaction.
NH₄Cl + AgNO₃ ⇒ AgCl + NH₄NO₃
2.- Calculate the molar mass of NH₄Cl and AgCl
NH₄Cl = 14 + 4 + 35.5 = 53.5 g
AgCl = 108 + 35.5 = 143.5 g
3.- Calculate the theoretical yield
53.5 g of NH₄Cl -------------------- 143.5 g of AgCl
15.5 g of NH₄Cl ------------------- x
x = (15.5 x 143.5) / 53.5
x = 2224.25 / 53.5
x = 41.6 g of AgCl