Answered

Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

if 315mL of 1.25 M iron(ii) sulfate is added to 405mL of.95 M ammonium phosphate, what is the limiting reactant


Sagot :

drpelezo

Answer:

FeSO₄ is the Limiting Reactant

Explanation:

3FeSO₄(aq) + 2(NH₄)₃PO₄ (aq) => 3(NH₄)₃PO₄(aq) + Fe₃(PO₄)(s)

315ml(1.25M)FeSO₄ + 405ml(0.95M)(NH₄)₃PO₄

=> 0.315L(1.25M)FeSO₄       +      0.405ml(0.95M)(NH₄)PO₄

=> 0.394mole FeSO₄           +      0.385mole (NH₄)PO₄

Divide by respective coefficient => the smaller value is the Limiting Reactant.

=> 0.394/3 = 0.131                      => 0.385/2 = 0.193

Smaller value is associated with FeSO₄

∴the limiting reactant is FeSO₄.

Summery:

1. convert reactant quantities to moles

2. divide reactants by respective coefficients of balanced equation

3. note smaller value after dividing by respective coefficient => Limiting Reactant.

Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.