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shtirl [24]
3 years ago
6

65.5g of copper reacts with __g of oxygen to produce 81g copper(I) oxide

Chemistry
1 answer:
Nikitich [7]3 years ago
7 0

Answer:

grams O₂(g) ≅ 15.9 g O₂(g) (3 sig. figs.)  

Explanation:

2Cu + O₂ => 2CuO

Convert given to moles, solve by equation rxn ratio then convert to grams.

Moles CuO formed = 81.0 g / 81.5 g·mol⁻¹ = 0.994 mol CuO

0.994 mol CuO requires 0.994 mol Cu = 0.994 mol Cu x 64 g·mol⁻¹ = 63.6 g Cu => from this, the amount of O₂ needed = 1/2(0.994 mol) O₂(g) = 0.497 mol O₂(g) = (0.497 mol O₂(g))(32 g·mol⁻¹) = 15.904 g O₂(g) ≅ 15.9 g O₂(g) (3 sig. figs.)                

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what is free radical halogenation?

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Bromination of an alkane includes the substitution of a bromine atom for a hydrogen atom. The following stages will be taken by 2,2,4-trimethylpentane during this reaction:

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Termination: The remaining free radical of bromide reacts with the tertiary free radical of 2,2,4-trimethylpentane to form 2-bromo-2,4,4-trimethylpentane.

the principal organic product formed by free-radical bromination of 2,2,4−trimethylpentane is 2-bromo-2,4,4-trimethylpentane.

To know more about free radical halogenation, check out:

brainly.com/question/13046867

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