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1 answer:
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Answer:
1.44 x 10²⁵ ions of Na⁺
Explanation:
Given parameters:
Mass of NaCl = 1.4kg = 1400g
Unknown:
Number of ions of sodium = ?
Solution:
The compound NaCl in ionic form can be written as;
NaCl → Na⁺ + Cl⁻
In 1 mole of NaCl we have 1 mole of sodium ions
Now, let us find the number of moles in NaCl;
Number of moles =
Molar mass of NaCl = 23 + 35.5 = 58.5g/mol
Number of moles = = 23.93mol
So;
Since 1 mole of NaCl gives 1 mole of Na⁺
In 23.93 mole of NaCl will give 23.93 mole of Na⁺
1 mole of a substance = 6.02 x 10²³ ions of a substance
23.93 mole of a substance = 6.02 x 10²³ x 23.93
= 1.44 x 10²⁵ ions of Na⁺
<u>Answer:</u> The percent yield of the reaction is 91.8 %
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For
:</u>
Given mass of = 4.0 g
Molar mass of = 63.12 g/mol
Putting values in equation 1, we get:
- <u>For oxygen gas:</u>
Given mass of oxygen gas = 10.0 g
Molar mass of oxygen gas = 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:
12 moles of oxygen gas reacts with 2 moles of
So, 0.3125 moles of oxygen gas will react with = of
As, given amount of is more than the required amount. So, it is considered as an excess reagent.
Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
12 moles of oxygen gas produces 5 moles of
So, 0.3125 moles of oxygen gas will produce = of water
Now, calculating the mass of from equation 1, we get:
Molar mass of = 69.93 g/mol
Moles of = 0.130 moles
Putting values in equation 1, we get:
To calculate the percentage yield of , we use the equation:
Experimental yield of = 8.32 g
Theoretical yield of = 9.052 g
Putting values in above equation, we get:
Hence, the percent yield of the reaction is 91.8 %