QUESTION 2: How much Pareactant will be needed to produce 105.8g of tetraphosphorus decaoxide (P4010)? G: 105.8g P4010 мM: 1 тol Pa010 — 287.88g Р,010 1 mol P4 = 127.88g P4 MR: 3 тol P,01о — Зтоl Р, СHOICES: A. 74g P4 B. 0.74g P4 С. 47g Ра D. 0.047g P4

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Chapter4: Stoichiometry
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QUESTION 2: How much Pareactant will be needed to produce 105.8g of tetraphosphorus decaoxide
(P4010)?
G: 105.8g P4010
мм: 1 тol P,010 3D 287.88g P,010
1 тol P, 3D 127.88g Р,
MR: 3 тol P,01о — Зтоl Р,
СHOICES:
A. 74g P4
В. 0.74g Ра
С. 47g Ра
D. 0.047g P4
Transcribed Image Text:QUESTION 2: How much Pareactant will be needed to produce 105.8g of tetraphosphorus decaoxide (P4010)? G: 105.8g P4010 мм: 1 тol P,010 3D 287.88g P,010 1 тol P, 3D 127.88g Р, MR: 3 тol P,01о — Зтоl Р, СHOICES: A. 74g P4 В. 0.74g Ра С. 47g Ра D. 0.047g P4
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