Answer:
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Explanation:
2 Atom of Hydrogen are present in 2 (NaHCO3)
The balanced chemical equation that illustrates this reaction is:
<span>C2H4 + 3O2 --> 2CO2 + 2H2O
</span>
From the periodic table:
mass of carbon = 12 grams
mass of hydrogen = 1 gram
Therefore:
molar mass of C2H4 = 12(2) + 4(1) = 24 + 4 = 28 grams
number of moles = mass / molar mass
number of moles of C2H4 = 54.7 / 28 = 1.95 moles
From the balanced equation above:
3 moles of oxygen are required to react with one mole of C2H4, therefore, to know the number of moles required to react with 1.95 moles of C2H4, all you have to do is cross multiplication as follows:
number of oxygen moles = (1.95*3) / 1 = 5.85 moles
Answer:
-3135.47 kJ/mol
Explanation:
Step 1: Write the balanced equation
C₆H₆(l) + 7.5 O₂(g) ⇒ 6 CO₂(g) + 3 H₂O(g)
Step 2: Calculate the standard enthalpy change of the reaction (ΔH°r)
We will use the following expression.
ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)
where,
n: moles
ΔH°f: standard enthalpies of formation
p: products
r: reactants
ΔH°r = 6 mol × ΔH°f(CO₂(g)) + 3 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C₆H₆(l)) - 7.5 mol × ΔH°f(O₂(g))
ΔH°r = 6 mol × (-393.51 kJ/mol) + 3 mol × (-241.82 kJ/mol) - 1 mol × (48.95 kJ/mol) - 7.5 mol × 0 kJ/mol
ΔH°r = -3135.47 kJ
Since this enthalpy change is for 1 mole of C₆H₆(l), we can express it as -3135.47 kJ/mol.
Answer:
=71.76 grams
Explanation:
At Room temperature and pressure, 1 mole of an ideal gas occupies a volume of 24 liters.
Therefore, the number of moles occupied by 28.7 L is:
(28.7×1)/24=1.196 moles.
Mass=Number of moles× RMM
RMM of N₂O₂ is 60
mass=1.196 moles× 60 grams/mol
=71.76 grams
