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The Answer to the question is 14 grams approximately.
Atomic mass of Na = 23 u
Given mass of Na = 3.4 gm
Moles of Na = given mass of Na / molar mass of Na
Moles of Na = 3.4 / 23 moles = 0.15 moles approximately
So, 0.15 moles of Na approximately
Molar mass of Cl₂ = 35.45 × 2 = 70.9 u
Given mass of Cl₂ = 8.9 gm
Moles of Cl₂ = given mass of Cl₂ / molar mass of Cl₂
Moles of Cl₂ = 8.9 / 70.9 moles = 0.12 moles approximately
So, 0.12 moles of Cl₂ approximately
Balanced chemical equation
2Na + Cl₂ → 2NaCl
Ratio of Moles to Stoichiometric coefficients
0.15/2 0.12/1
0.075 < 0.12
So, as ratio of Na is less then it will be our Limiting reagent.
Which means that Cl₂ will consume complete in the reaction.
It is clear from the reaction that 1 moles of Cl₂ are making 2 mole of NaCl
1 moles of Cl₂ → 2 mole of NaCl
0.12 moles of Cl₂ → 2 × 0.12 mole of NaCl
0.12 moles of Cl₂ → 0.24 mole of NaCl
So, 0.12 moles of Cl₂ will make 0.24 mole of NaCl
0.24 mole NaCl = mass formed NaCl / molar mass of NaCl
mass formed NaCl = 0.24 × molar mass of NaCl
mass formed NaCl = 0.24 × 58.44
mass formed NaCl = 0.24 × 58.44
mass formed NaCl = 14 grams (approximately)
14 grams of NaCl will be formed.
So, approximately 14 grams NaCl is formed from 3.4 grams of Na and 8.9 grams of Cl.
Learn more about Limiting Reagent here:
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