Answer:
ΔH=15000
J = 15KJ
Explanation:
In this exercise you have find the enthalpy of reaction this is the difference between enthalpy of reactans and products,
For the following equation
H2A(aq) + 2 BOH(aq) → B2A(aq) + 2 H2O(l)
We know that 0.20 moles of BOH reacted with excess amount of H2A solution and 1500. J
so,
(2mol/0,2mol)*1500J=15000J
for de reactions exothermics tha enthalpy is negative so:
ΔH=15000
J = 15KJ
Solids: wood, plastic, metals
Liquids: water (H2O), soap, juice, milk
Gases: Air, water vapour (H2O), carbon dioxide (CO2)
An element: gold, silver, aluminum, copper, nickel, iron, nitrogen in the air
A compound: salt (NaCl), carbon dioxide (CO2)
A homogenous mixture: air, water with sugar or salt dissolved in it
A heterogeneous mixture: orange juice with pulp, cereal in milk
Pure substances: diamond, water, salt, baking soda, sugar
Answer:
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Answer:
Oxygen is the limiting reactant.
Explanation:
Hello,
In this case, given the reaction:
Hence, given the masses of both ethanol and oxygen, we are able to compute the available moles ethanol by:
Next, we compute the moles of ethanol that react with the 0.640 grams of oxygen considering their 1:3 molar ratio in the chemical reaction:
In such a way, since there are 0.01 available moles of ethanol but just 0.0067 moles are reacting, we evidence ethanol is in excess, therefore the oxygen is the limiting reactant.
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