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Sagot :
The percent yield of H₂O, if 87.0 g of H₂O is produced by combining 95.0 g of O₂ and 11.0 g of H₂ is 87.87%.
How do we calculate mass from moles?
Mass of any substance will be calculated by using their moles as:
n = W/M, where
W = given or required mass
M = molar mass
Moles of 95g of Oxygen (O₂) = 95g / 32g/mol = 2.96 moles
Moles of 11g of hydrogen (H₂) = 11g / 2g/mol = 5.5 moles
Given chemical reaction is:
2H₂ + O₂ → 2H₂O
From the stoichiometry of the reaction, it is clear that:
1 moles of O₂ = reacts with 2 moles of H₂
2.96 moles of O₂ = reacts with 2×2.96=5.92 moles of H₂
Here hydrogen is the limiting reagent as it has lower moles and formation of water depends on this only.
2 moles of H₂ = produces 2 moles of water
5.5 moles of H₂ = produces 5.5 moles of water
Mass of 5.5 moles of water will be calculated as:
W = (5.5mol)(18g/mol) = 99g
Given theoretical yield of water = 87g
% yield of water will be calculated as:
% yield = (87 / 99)×100 = 87.87%
Hence required value is 87.87%.
To know more about % yield, visit the below link:
https://brainly.com/question/25996347
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