Answer:
Hydride ion
Explanation:
You often see the reduction with NADH written as
NADH ⟶ NAD⁺ + H⁺ + 2e⁻
If you think about it, H⁺ + 2e⁻ is equivalent to H:⁻, so we could write the reaction as
NADH ⟶ NAD⁺ + H:⁻
In terms of a mechanism, the dihydropyridine ring of NADH transfers a hydrogen atom with its pair of electrons (a hydride ion) to the substrate and becomes the more stable, aromatic pyridinium ion in NAD⁺.
The volume of the given solution is 2.5 L.
<u>Explanation:</u>
Volume of the solution = 2500 mL
We have to convert it into litres as,
1000 mL = 1 litre
To convert ml into litres we have to divide the millilitres by 1000, so that it can be converted into L.
Here given volume is 2500 mL.
Volume in L =
= 2.5 L
So the volume of the solution is 2.5 L.
<u>Answer:</u> The mass of iron (III) oxide produced is 782.5 grams
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of = 588 g
Molar mass of = 120 g/mol
Putting values in equation 1, we get:
Given mass of = 352 g
Molar mass of = 32 g/mol
Putting values in equation 1, we get:
The chemical equation for the reaction of and oxygen gas follows:
By Stoichiometry of the reaction:
1 mole of reacts with 1 mole of oxygen gas
So, 4.9 moles of will react with =
of oxygen gas
As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.
Thus, is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of produces 1 mole of iron (III) oxide
So, 4.9 moles of will produce =
of iron (III) oxide
Now, calculating the mass of iron (III) oxide from equation 1, we get:
Molar mass of iron (III) oxide = 159.7 g/mol
Moles of iron (III) oxide = 4.9 moles
Putting values in equation 1, we get:
Hence, the mass of iron (III) oxide produced is 782.5 grams