Answer: 16.32 g of as excess reagent are left.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of require = 1 mole of
Thus 0.34 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and
is the excess reagent.
Moles of left = (0.68-0.17) mol = 0.51 mol
Mass of
Thus 16.32 g of as excess reagent are left.
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.
= 2*(MM of Cr) + 3*( MM of S) + 3*4*( MM of O)
= 2*(52) + 3*(32) + 12*(16)
= 104 + 96 + 192
= 392 g
= 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.