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How many grams of CIF3 form when .204 moles of F2 react with excess Cl2? The answer must be sold with dimensional analysis.
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Mar 27, 2017
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How many grams of CIF3 form when .204 moles of F2 react with excess Cl2?
The answer must be sold with dimensional analysis.
Chemistry
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Dieter B
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The reaction equation:
2Cl₂ + 3F₂ → 2ClF₃
Moles of F₂ = 0.204
Moles of ClF₃ = 2/3 x 0.204
= 0.136 mole
Mass = moles x Mr
= 0.136 x (35.5 + 19 x 3)
= 12.58 grams of ClF₃ will be formed.
Kbrown
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Mar 31, 2017
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