The number of moles of oxygen required to generate 28 moles of water from the reaction is 14 moles
<h3>Balanced equation </h3>
2H₂ + O₂ —> 2H₂O
From the balanced equation above,
2 moles of water were obtained from 1 mole of oxygen
<h3>How to determine the mole of oxygen needed </h3>
From the balanced equation above,
2 moles of water were obtained from 1 mole of oxygen
Therefore,
28 moles of water will be obtained from = 28 / 2 = 14 moles of oxygen
Thus, 14 moles of oxygen are needed for the reaction
Learn more about stoichiometry:
brainly.com/question/14735801
A.) Delta . Delta is a landform that forms from Deposition.
Answer:
When nitric acid combine with sodium hydroxide the salt formed is called sodium nitrate. option B
Explanation:
It is the strong acid strong base reaction. When acid and base react with each other salt and water are formed.
In given reaction nitric acid combine with sodium hydroxide base and form sodium nitrate salt and water.
Chemical equation:
HNO₃(aq) + NaOH(aq) → NaNO₃(aq) + H₂O(l)
Ionic equation:
H⁺NO₃⁻(aq) + Na⁺OH⁻(aq) → Na⁺NO₃⁻(aq) + H₂O(l)
Net ionic equation:
H⁺(aq) + OH⁻(aq) → H₂O(l)
The Na⁺(aq) and NO₃⁻(aq) are spectator ions that's why these are not written in net ionic equation. The water can not be splitted into ions because it is present in liquid form.
Spectator ions:
These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.
Answer:
163.2g
Explanation:
First let us generate a balanced equation for the reaction. This is shown below:
4Al + 3O2 —> 2Al2O3
From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.
From the equation,
4moles of Al produced 2moles of Al2O3.
Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.
Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:
Mole of Al2O3 = 1.6mole
Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol
Mass of Al2O3 =?
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass of Al2O3 = 1.6 x 102 = 163.2g
Therefore the theoretical of Al2O3 is 163.2g
