Answer:
solid substance formed during liquid solution, is known as precipitate. For example: in the reaction of silver nitrate and sodium chloride, a precipitate of silver chloride is formed which is white in colour. Precipitation can be of any colour like white, yellow, black.
For Anyone that put A, its wrong, the correct answer is D
When 1 mole of Zn and 2 mole of O₂ reacts togethor, It will produce 1 mole of ZnO as O₂ is excess reagent and Zn will act as Limiting reagent and thus it limits the amount of product formed.
<h3>What is Limiting reagent ?</h3>
Limiting reagents are the substances that are completely consumed first in a chemical reaction.
Given ;
- Amopunt of Zn : 1 mole
- Amount of O₂ : 2 mole
Given equation ;
2Zn + O₂ --> 2ZnO
As according to given chemical equation,
2 moles Zn of reacts with 1 mole of O₂ to produce 2 moles of ZnO.
Therefore,
1 mole of Zn will require 0.5 mole of O₂.
But,
the given amount of O₂ is 2 mole which is excess
Hence,
Zn here will act as the Limiting reagent, According to which the amount of product formed will be decided.
Therefore,
If 2 mole Zn produces : 2 moles of ZnO (According to given balanced equation)
Thus,
1 mole of Zn will produce = 2/2 x 1
= 1 mole of ZnO
Hence, When 1 mole of Zn and 2 mole of O₂ reacts togethor, It will produce 1 mole of ZnO as O₂ is excess reagent and Zn will act as Limiting reagent and thus it limits the amount of product formed.
Learn more about Limiting reagent here ;
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According to the balanced equation of the reaction:
H2X (aq) +2 NaOH (aq) → Na2X (aq) + 2H2O(l)
First, we have to get the no. of moles of H2X:
no.of moles of H2X = weight / molar mass
when we have the H2X weight = 0.1873 g & the molar mass H2X = 85 g/mol
So by substitution:
∴ no.of moles of H2X = 0.1873 /85
= 0.0022 mol
-then, we need to get no.of moles of NaOH:
from the balanced equation, we can see that 1 mol H2X = 2 mol NaOH
∴ no.of moles of NaOH = no.of moles of H2X *2
= 0.0022 * 2 = 0.0044 mol
So we can get the volume per litre from this formula:
M (NaOH) = no.of moles NaOH / Volume L
So by substitution:
0.1052 = 0.0044 / Volume L
∴Volume = 0.042 L *1000 = 42 mL
