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Olin [163]
3 years ago
14

Combustion of hydrogen releases 142 j/g of hydrogen reacted. How many kj of energy are released by the combustion of 16.0 oz of

hydrogen? (1 lb = 16 oz; 1 kg = 2.2 lb)

Chemistry
2 answers:
Delvig [45]3 years ago
7 0

64.61 kJ of energy is released by the combustion of 16.0 oz of hydrogen

<h3>Further explanation</h3>

Delta H reaction (ΔH) is the amount of heat/heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The reaction of a substance with oxygen is called a combustion reaction

the combustion of hydrogen is

<h3>2H₂ + O₂ → 2H₂O.</h3>

This equation shows that 2 Hydrogen molecules react with 1 molecule of oxygen to produce 2 molecules of water

In this combustion reaction will produce large thermal energy because it involves breaking the hydrogen gas bond, so the reaction is exothermic

The combustion of hydrogen releases 142 j / g of hydrogen reacted.

Combustion of 16.0 oz of hydrogen will release heat of:

16 oz = 1 lb = 0.455 kg = 455 g

For each gram Hydrogen releases 142 J, then for 455 g releases:

142 x 455 J = 64610 J = 64.61 kJ

<h3>Learn more</h3>

the complete combustion of methane

brainly.com/question/1971314

grams of o2 are required to burn 17.0 gal of c8h18

brainly.com/question/5816411#

moles of oxygen are required to produce 2.33 moles of water

brainly.com/question/6209439

Keywords: Combustion of hydrogen, ΔH reaction, water, oxygen, exothermic

Andrej [43]3 years ago
5 0

Given the mass of hydrogen = 16.0 oz

Converting 16.0 oz hydrogen to pounds (lb) using the conversion factor 1 lb = 16 oz:

16.0 oz * \frac{1 lb}{16 oz} =1 lb

Converting 16.0 lb to g using the conversion factors 1 kg = 2.2 lb, 1 kg = 1000 g:

1lb * \frac{1kg}{2.2lb}*\frac{1000g}{1kg}= 454.5 g

Heat of combustion of hydrogen = 142 J/g

Calculating the heat released when 16.0 oz is combusted:

454.5g H_{2} * \frac{142 J}{g} *\frac{1 kJ}{1000J}=64.5kJ


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<u>________________________________________</u>

<u>Gram</u><u> </u><u>Molecular</u><u> </u><u>Mass</u><u> </u><u>of</u><u> </u><u>Propane</u><u>(</u><u>C3H8</u><u>)</u>

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\mathsf{ \frac{44 }{6.02 \times  {10}^{23} } \times 4.75 \times  {10}^{23}  }

\mathsf{  = \frac{44 }{6.02 \times   \cancel{{10}^{23} }} \times 4.75 \times \cancel{ {10}^{23}}  }

\mathsf{  = \frac{44 }{6.02 } \times 4.75   }

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