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

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

General, Organic, and Biological Chemistry

7th Edition

ISBN:9781285853918

Author:H. Stephen Stoker

Publisher:H. Stephen Stoker

Chapter6: Chemical Calculations: Formula Masses, Moles, And Chemical Equations

Section: Chapter Questions

Problem 6.68EP: How many carbon monoxide molecules (CO) are needed to react with 8 hydrogen molecules (H2) to...

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