What is the mass in grams of 2.5 mol of O2?
1 answer:
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Taking into account the reaction stoichiometry, 10.93 grams of CO₂ are formed when 8.50 g of methane react with 15.9 g of O₂.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
CH₄ + 2 O₂ → CO₂ + 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- CH₄: 1 mole
- O₂: 2 moles
- CO₂: 1 mole
- H₂O: 2 moles
The molar mass of the compounds is:
- CH₄: 16 g/mole
- O₂: 32 g/mole
- CO₂: 44 g/mole
- H₂O: 18 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- CH₄: 1 mole ×16 g/mole= 16 grams
- O₂: 2 moles ×32 g/mole= 64 grams
- CO₂: 1 mole ×44 g/mole= 44 grams
- H₂O: 2 moles ×18 g/mole=36 grams
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
<h3>Limiting reagent in this case</h3>To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 16 grams of CH₄ reacts with 64 grams of O₂, 8.50 grams of CH₄ reacts with how much mass of O₂?
mass of O₂= 34 grams
But 34 grams of O₂ are not available, 15.9 grams are available. Since you have less mass than you need to react with 8.50 grams of CH₄, O₂ will be the limiting reagent.
<h3>Mass of CO₂ formed</h3>The following rule of three can be applied: if by reaction stoichiometry 64 grams of O₂ form 44 grams of CO₂, 15.9 grams of O₂ form how much mass of CO₂?
<u><em>mass of CO₂= 10.93 grams</em></u>
Then, 10.93 grams of CO₂ are formed when 8.50 g of methane react with 15.9 g of O₂.
Learn more about the reaction stoichiometry:
#SPJ1
4C₃H₅(NO₃)₃ ------> 12CO₂
+ 6N₂
+ 10H₂O
+ O₂
mol of CO₂ =
=
mol ratio of CO₂ : C₃H₅(NO₃)₃
12 : 4
∴ if mole of CO₂ = 0.568 mol
then " " C₃H₅(NO₃)₃ =
= 0.189 mol
∴ mass of nitroglycerin = mole * Mr
= 0.189 mol * 227.0995 g / mol
= 43.00 g
Answer:
i)The carbon particles are formed when the gas is not completely burnt(True)
ii)A solid or liquid must be change to gaseous state in order to make flame(TRUE)
iii)A non- luminous Bunsen flame produces has three visible regions(FALSE)
iv)The luminous Bunsen flame produces less soot than the non- luminous flame(FALSE)
v)A yellow Bunsen flame is obtained when the air holes are opened(FALSE)
vi)The luminous Bunsen flame is hotter than a non- luminous Bunsen flame(FALSE)
vii)Bunsen burner uses kerosene as a source of heat(FALSE)
Explanation:
i)When the incomplete combustion ,then CO(carbon monoxide) and C(carbon) are produced. <u>The unburnt carbon get deposited as</u><u> Soot</u>.
In complete combustion CO2(carbon dioxide) is produced.
ii)Fire, is the mixture of hot gases.Flame are result of chemical reaction between oxygen and fuel(e.g butane ,wood).Flame contain,carbon dioxide , heat ,light,steam.
iii)There are total three regions of flame in which the outer most region is the non-luminous .The non-luminous part is <u>blue(one visible region</u>) and does not emit-light.
iv) Luminous Bunsen flame produces more soot than the non- luminous flame because luminous flame contain more unburnt particles as compare to non-luminous.
luminous part has less oxygen available for complete combustion of fuel
v)When the air hole is open it means excess oxygen is supplied to the fuel.
In this case clean and luminous flame is produced.When air hole is close yellow flame is produced.
vi)Non-luminous part of flame is more hot because of complete combustion due to proper availability of oxygen. Sometime the particle of non-luminous flame get ionize also.
vii).The bunsen burner uses Natural gas(methane) or LPG(propane and butane)as sources of heat because these are easily burnt not kerosene.
