How many moles of Al are necessary to form 45.6 g of AlBr, from this reaction: 2 Al(s) + 3 Br₂(1)→ 2 AlBr,(s)?

Chemistry & Chemical Reactivity
9th Edition
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter3: Chemical Reactions
Section: Chapter Questions
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How many moles of Al are necessary to form 45.6 g of AIBr, from this
reaction:
2 Al(s) + 3 Br₂(1)→ 2 AlBr₂(s)?
Transcribed Image Text:How many moles of Al are necessary to form 45.6 g of AIBr, from this reaction: 2 Al(s) + 3 Br₂(1)→ 2 AlBr₂(s)?
Starting with the following equation,
Pb(NO₂)₂(aq) + K₂PO₂(aq) → Pb₂(PO)(s) + KNO₂(aq)
calculate the moles of Pb(NO3), that will be required to produce 885
grams of Pb (PO₂)₂.
Transcribed Image Text:Starting with the following equation, Pb(NO₂)₂(aq) + K₂PO₂(aq) → Pb₂(PO)(s) + KNO₂(aq) calculate the moles of Pb(NO3), that will be required to produce 885 grams of Pb (PO₂)₂.
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