2 C4H10 + 13 02 → 8 CO2 + 10 H2OWhich of the following is the correct set up to solve for how many grams of oxygenare needed to react with 16.4 g of butane, C4H10?16.4 g C,H101 mol CH102 mol CH1032.00 g 0258.14 g C,H10 | 13 mol O,1 mol O216.4 g C,H,01 mol CH1013 mol O,32.00 g O258.14 g C,H10 2 mol C,H101 mol O21 mol CH108 mol 0232.00 g 0216.4 g CH1058.14 g C,H10 2 mol C,H101 mol O216.4 g CH101 mol CH1013 mol O216.00 g O258.14 g C,H10| 2 mol CH101 mol O2

Given :- mass of butane = 16.4 g To calculate :- mass of O2 required number of moles = mass/molar…

Answered: Which of the following is the correct…

Which of the following is the correct set up to solve for how many grams of oxygen are needed to react with 16.4 g of butane, C4H10?

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter3: Mass Relations In Chemistry; Stoichiometry

Section: Chapter Questions

Problem 55QAP: One way to remove nitrogen oxide (NO) from smoke-stack emissions is to react it with ammonia....

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Concept explainers

Stoichiometry

Stoichiometry is a chemical division that evaluates the quantitative content of the chemical process with the help of the reacting molecules or products taking part in the reaction. The term “stoicheion” in “Stoichiometry”, is an old Greek term for "element" whereas “metron” means "measure".

Question

2 C4H10 + 13 02 → 8 CO2 + 10 H2O
Which of the following is the correct set up to solve for how many grams of oxygen
are needed to react with 16.4 g of butane, C4H10?
16.4 g C,H10
1 mol CH10
2 mol CH10
32.00 g 02
58.14 g C,H10 | 13 mol O,
1 mol O2
16.4 g C,H,0
1 mol CH10
13 mol O,
32.00 g O2
58.14 g C,H10 2 mol C,H10
1 mol O2
1 mol CH10
8 mol 02
32.00 g 02
16.4 g CH10
58.14 g C,H10 2 mol C,H10
1 mol O2
16.4 g CH10
1 mol CH10
13 mol O2
16.00 g O2
58.14 g C,H10| 2 mol CH10
1 mol O2

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