Question

A sample of 85.5 g of tetraphosphorous decoxide (P4O10) reacts with 74.9 g of water to produce phosphoric acid (H3PO4) according to the following balanced equation.
P4O10+6H2O⟶4H3PO4

Determine the limiting reactant for the reaction.

H2O

H3PO4

P4O10
Calculate the mass of H3PO4 produced in the reaction.
mass of H3PO4:
g
Calculate the percent yield of H3PO4 if 39.2 g of H3PO4 is isolated after carrying out the reaction.
percent yield:
%

Answer 1

The limiting reactant here is P4O10 . The percent yield of the product is caculated as 33.3%.

What is the limiting reactant?

The limiting  reactant is the reactant that is in a minute quantity.

Number of moles of P4O10 =  85.5 g/284 g/mol = 0.3 moles

Number of moles of H2O =  74.9 g / 18 g/mol = 4.2 moles

From the reaction equation;

1 mole of P4O10 reacts with 6 moles of H2O

x moles of P4O10 reacts with 4.2 moles of H2O

x = 0.7  moles

Hence, P4O10 is the limiting reactant.

1 mole of P4O10 yields 4 moles of H3PO4

0.3 moles of P4O10 yields

0.3 moles * 4/1 = 1.2 moles

Mass of the H3PO4 = 1.2 moles * 98 g/mol = 117.6 g

Percent yield = 39.2 g/117.6 g * 100/1 = 33.3 %

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