The average weight of an atom of an element, formerly based on the
weight of one hydrogen atom taken as a unit or on 1/16 (0.0625) the
weight of an oxygen atom, but after 1961 based on 1/12 the weight of the
carbon-12 atom.
Answer:
= 3.78 g H₂O
Explanation:
2C₂H₆ + 3O₂ => 4CO₂ + 6H₂O
2.1g C₂H₆ = 2.1g/30.0 g/mol = 0.07 mole ethane
3.68g O₂ = 3.68g/32 g/mol = 0.115 mole oxygen
Limiting Reactant:
A quick way to determine limiting reactant is to divide moles of reactant by its respective coefficient in the balanced molecular equation. The smaller value is the limiting reactant.
moles ethane = 0.07 mole / 2 (the coefficient in balanced equation) = 0.035
moles oxygen = 0.115 mole / 3 (the coefficient in balanced equation) = 0.038
Since the smaller value is associated with ethane, then ethane is the limiting reactant and the problem is worked from the 0.07 moles of ethane in an excess of O₂.
From the equation stoichiometry ...
2 moles C₂H₆ in an excess of O₂ => 6 moles H₂O
then 0.07 mole C₂H₆ in an excess of O₂ => 6/2(0.07 moles H₂O = 0.21 mole
Converting to grams of water produced
= 0.21 mole H₂O X 18 g/mol = 3.78 g H₂O
Answer: Correct option is B)
The standard enthalpy change of formation of a compound is the enthalpy change which occurs when one mole of the compound is formed from its elements under standard conditions. The equation showing the standard enthalpy change of formation of water is
H
2
(g)+
2
1
O
2
(g)⟶H
2 ↑
O(l) 1 mole of water formed.
∴ Enthalpy of formation is −X2kJ/mol.
________________________________________________________
Explanation:I hope this helped
Answer:
Yes, that is correctly balanced. 2 H2 + 1 O2 = 2 H2O is already balanced.
Explanation:
If you count up the number of hydrogen atoms and oxygen atoms on each side, you will see that they are equal:
2 H2 + 1 O2 (4 hydrogen and 2 oxygen) = 2 H2O (4 hydrogen and 2 oxygen).
