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For the reaction 2A 3B 5C---3 Initially if 2 mole of A, 4mole of B and 6 mole of C is taken, with 25percent yield, moles of D which can be produced are?
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For the reaction 2A 3B 5C---3 Initially if 2 mole of A, 4mole of B and...
Calculation of moles of reactants
- Given that 2 moles of A, 4 moles of B, and 6 moles of C are taken initially.
- Therefore, the number of moles of reactants is:
- Moles of A = 2
- Moles of B = 4
- Moles of C = 6

Calculation of limiting reactant
- To determine the limiting reactant, we need to calculate the moles of product formed by each of the reactants.
- The balanced chemical equation is 2A + 3B + 5C → 3D
- The moles of product formed by each of the reactants are:
- Moles of D formed by 2 moles of A = 3/2 moles
- Moles of D formed by 4 moles of B = 2 moles
- Moles of D formed by 6 moles of C = 3.6 moles
- Since the moles of D formed by A are the least, A is the limiting reactant.

Calculation of moles of product formed
- Given that the yield is 25%
- Therefore, the moles of D formed are:
- Moles of D = 3/2 * 0.25 = 0.375 moles

Conclusion
- The moles of D that can be produced are 0.375 moles.
Community Answer
For the reaction 2A 3B 5C---3 Initially if 2 mole of A, 4mole of B and...
Is 3/4 the ans?
A is the limiting reagent
2 moles of a gives 3 moles of D
%=25
X/3×100=25% (X moles of D )
solving this we get moles of D are equal to
3/4
cross checking
----->moles formed =3/4 should form =3
%= (3/4×3)×100 =25%
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For the reaction 2A 3B 5C---3 Initially if 2 mole of A, 4mole of B and 6 mole of C is taken, with 25percent yield, moles of D which can be produced are?
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