PLEASE HELP!! ASAP WILL GIVE BRAILIST (please explain how you got there or show a picture of work) a) Write the word equation for the

Question

PLEASE HELP!! ASAP WILL GIVE BRAILIST (please explain how you got there or show a picture of work)
a) Write the word equation for the reaction of silver chlorate reacts with sodium phosphate to produce silver phosphate and sodium chlorate.
b) Write the balanced formula equation for this reaction.
c) If 287 g of silver chlorate reacts with 52.8 g of sodium phosphate, what is the maximum number of grams mass of silver phosphate will be produced?
d) Identify which reactant is the limiting reactant and which is the excess reactant.
e) How many grams of the excess reactant reacts and how many grams are left over?

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Ngọc Khuê 3 years 2021-08-26T02:46:37+00:00 1 Answers 11 views 0

Answers ( )

  1. thugiang
    0
    2021-08-26T02:47:41+00:00
    Reply

    Answer:

    See Explanation

    Explanation:

    a) The word equation is;

    silver chlorate + sodium phosphate ——> silver phosphate + sodium chlorate

    b) The balanced reaction equation of the reaction is;

    3AgClO3(aq) + Na3PO4(aq) ——->Ag3PO4(s) + 3NaClO3(aq)

    c) amount of silver chlorate reacted = 287 g/191.319 g/mol = 1.5 moles

    3 moles of silver chlorate yields 1 mole of silver phosphate

    1.5 moles of silver chlorate yields 1.5 * 1/3 = 0.5 moles of silver phosphate

    Amount of sodium phosphate reacted = 52.8 g/163.94 g/mol = 0.3 moles

    1 mole of sodium phosphate yields 1 mole of silver phosphate

    0.3 moles of sodium phosphate yields 0.3 moles of silver phosphate.

    Hence the maximum number of grams mass of silver phosphate produced = 0.3 moles * 418.58 g/mol = 125.6 g

    d) The excess reactant is silver chlorate while the limiting reactant is sodium phosphate

    e) 3 moles of silver chlorate  reacts with 1 mole of sodium phosphate

      x moles of silver chlorate reacts with 0.3 moles of sodium phosphate

    x = 3 * 0.3 = 0.9 moles of silver chlorate reacted

    Mass of 0.9 moles of silver chlorate = 0.9 moles * 191.319 g/mol  = 172 g of silver chlorate

    Hence mass of silver chlorate left over =  287 g – 172 g = 115 g

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