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Answer:
600 g K₂SO₄
Explanation:
First write down the complete, balanced chemical equation for such question. In this case:
Cr₂(SO₄)₃ + 2K₃PO₄ → 3K₂SO₄ + 2CrPO₄
Next, calculate molar masses of required compounds mentioned in the question. In this case it is for Cr₂(SO₄)₃ and K₂SO₄.
- Molar mass(MM) of Cr₂(SO₄)₃:
= 2*(MM of Cr) + 3*( MM of S) + 3*4*( MM of O)
= 2*(52) + 3*(32) + 12*(16)
= 104 + 96 + 192
= 392 g
- Molar mass(MM) of K₂SO₄:
= 2*(MM of K) + 1*( MM of S) + 4*( MM of O)
= 2*(39) + 32 + 4*(16)
= 78 + 32 + 64
= 174 g
Here comes the concept of Limiting reagent:
The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot continue without it. The other reactants present with this are usually in excess and called excess reactants. If quantities of both the reactants are given, then one should apply unitary method and find out the limiting reagent out of the two. Then, determine the amount of product formed or percentage yield.
Also, 1 mole( 392 g) of Cr₂(SO₄)₃ gives 3 moles( 174*3 = 522 g) of K₂SO₄.
Using unitary method, if 392g of Cr₂(SO₄)₃ gives 522 g of K₂SO₄ , then 450 g of Cr₂(SO₄)₃ will give how much of K₂SO₄?
Yeild of K₂SO₄ :
That is 599.3 g.
Since we have not considered molecular masses of individual atoms to 6 decimal places, this number can be approximated to 600g.
Therefore, 600g of K₂SO₄ is produced.
Explanation:
The given data is as follows.
Volume of water = 0.25
Density of water = 1000
Therefore, mass of water = Density × Volume
=
= 250 kg
Initial Temperature of water () =
Final temperature of water =
Heat of vaporization of water () at
is 2133 kJ/kg
Specific heat capacity of water = 4.184 kJ/kg/K
As 25% of water got evaporated at its boiling point () in 60 min.
Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg
Heat required to evaporate = Amount of water evapotaed × Heat of vaporization
= 62.5 (kg) × 2133 (kJ/kg)
= kJ
All this heat was supplied in 60 min = 60(min) × 60(sec/min) = 3600 sec
Therefore, heat supplied per unit time = Heat required/time = = 37 kJ/s or kW
The power rating of electric heating element is 37 kW.
Hence, heat required to raise the temperature from to
of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)
= 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)
= 125520 kJ
Time required = Heat required / Power rating
=
= 3392 sec
Time required to raise the temperature from to
of 0.25
water is calculated as follows.
= 56 min
Thus, we can conclude that the time required to raise the temperature is 56 min.