The combustion of isooctane (C₈H₁₈) is written below:
2 C₈H₁₈ (l) + 25 O₂ (g) → 16 CO₂ (g) + 18 H₂O (l)
The formula for heat of combustion is:
ΔHc = (∑Stoichiometric coefficient×ΔHf of products) - (∑Stoichiometric coefficient×ΔHf of reactants), where ΔHf is heat of formation.
ΔHf of isooctane = -259.2 kJ/mol
ΔHf of O₂ = 0 kJ/mol
ΔHf of CO₂ = -393.5 kJ/mol
ΔHf of H₂O = <span>-285.8 kJ/mol
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ΔHc = [(16 mol×-393.5 kJ/mol )+(18 mol×-285.8 kJ/mol)] - [(2 mol×-259.2 kJ/mol) + (25 mol*0 kJ/mol)]
ΔHc = -10,922 kJ
Answer:
moles of water in 1 ml = 1/18 mol.
take 1 ml and calculate it by 30
1 ml x 30= 30 ml
Explanation:
Answer:
Mass% Cr = 85.5%
Explanation:
<u>Given:</u>
Mass of CrBr3 sample = 0.8409 g
Mass of the AgBr precipitate = 1.0638 g
<u>To determine:</u>
The mass percent of Cr in the sample
<u>Calculation:</u>
The reaction of CrBr3 with silver nitrate results in the precipitation of the bromide ion as silver chloride (AgBr) and Cr as soluble Cr(NO3)2
CrBr3(aq) + 3AgNO3(aq)→ 3AgBr(s) + Cr(NO3)3(aq)
Molecular weight of AgBr =187.77 g/mol
Moles of AgBr precipitated is:
Since 1 mole of AgBr contains 1 mole of Cl, therefore:
# moles of Cl = 0.004478 moles
At wt of Cl = 35.45 g/mol
Mass%(Cr) = 100 - 14.50=85.5%
Answer:
When one reactant is in excess, there will always be some left over. The other reactant becomes a limiting factor and controls how much of each product is produced. While using excess reactants can help to increase percentage yields, this is at the expense of atom economy.
Explanation: