Which needs two less valence electrons to complete an octet, X+2?
1 answer:
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The image is not given in the question, it is attached below:
<u>Answer:</u> The excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.
<u>Explanation:</u>
In the given image:
Red spheres represent oxygen atoms, blue spheres represent nitrogen atoms and black spheres represent carbon atoms
The combination of 1 black and 2 red spheres will represent carbon dioxide compound
The combination of 2 blue spheres will represent nitrogen molecule
The combination of 1 blue and 1 red sphere will represent nitrogen monoxide compound
The combination of 1 black and 1 red sphere will represent nitrogen monoxide compound
Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.
Excess reagent is defined as the reagent which is left behind after the completion of the reaction.
We are given:
Given moles of NO = 6 moles
Given moles of CO = 4 moles
For the given chemical equation:
By stoichiometry of the reaction:
If 2 moles of CO reacts with 2 moles of NO
So, 4 moles of CO will react with = of NO
As the given amount of NO is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.
Thus, CO is considered a limiting reagent because it limits the formation of the product.
Hence, the excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.
Answer:
Formula: Na2S2O3
we get solubility.
Divide the mass of the compound by the mass of the solvent and then multiply by 100 g to calculate the solubility in g/100g .
Solution given:
mass of sodium thiosulphate [m1]=25.5g
mass of water [m2]=40g
at temperature [t]=25°C
we have
<u>solubility in g/dm^3</u> :
- =
- =63.75g /litre=63.75g/dm³
<u>solubility in g/dm^3 :63.75g/dm³</u>
<u>n</u><u>o</u><u>w</u>
solubility of the solute in mol/dm^3=:63.75g/dm³/178=0.4 mol/dm³