How much heat is required to increase the temperature of 198.5 grams of water from 25.0
1 answer:
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Answer: is the limiting reagent
Explanation:
To calculate the moles :
The balanced chemical reaction is :
According to stoichiometry :
1 moles of require = 2 moles of
Thus 0.625 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and
is the excess reagent as it is present more than the required amount.
The amount of heat required to convert H₂O to steam is : 382.62 kJ
<u>Given data :</u>
Mass of liquid water ( m ) = 150 g
Temperature of liquid water = 43.5°C
Temperature of steam = 130°C
<h3 /><h3>Determine the amount of heat required </h3>The amount of heat required = ∑ q1 + q2 + q3 ----- ( 1 )
where ;
q1 = heat required to change Temperature of water from 43.5°C to 100°C . q2 = heat required to change liquid water at 100°C to steam at 100°C
q3 = heat required to change temperature of steam at 100°C to 130°C
- For q1
M* S*ΔT
= 150 * 4.18 * ( 100 - 43.5 )
= 35425.5 J
- For q2
moles * ΔHvap
= (150 / 18 )* 40.67 * 1000
= 338916.67 J
- For q3
M * S * ΔT
= 150 * 1.84 * ( 130 -100 )
= 8280 J
Back to equation ( 1 )
Amount of heat required = 35425.5 + 338916.67 + 8280 = 382622.17 J
≈ 382.62 kJ
Hence we can conclude that The amount of heat required to convert H₂O to steam is : 382.62 kJ.
Learn more about Specific heat of water : brainly.com/question/16559442