Question

1. Given the following equation:

Answer 1

The percent yield obtained from the stoichiometry of the reaction when the moles of the reactants and products are known.

What is stoichiometry?

Stochiometry gives the relationship between the mass and mole or moles and volume in a reaction.

Q1)

K2PtCl4 + 2NH3 ----> Pt(NH3)2Cl2 +2KCl

Number of moles of NH3 = 34.5 g/17 g/mol = 2.03 moles

2 moles of NH3 yields 2 moles of KCl

2.03 moles  of NH3 yields 2.03 moles  of KCl

Theoretical yield of KCl =  2.03 moles * 75 g/mol = 154 g

Again,

2moles of NH3 yields 1 mole of Pt(NH3)2Cl2

2.03 moles  of NH3 yields 2.03 moles  * 1/2 = 1.015 moles

Mass of Pt(NH3)2Cl2 = 1.015 moles * 301 g/mol = 306 g

% yield =  76.4 g/306 g * 100/1 = 25%

Q2)

H3PO4 + 3 KOH ------> K3PO4 + 3 H2O

Number of moles of H3PO4  = 49.0 g/98 g/mol = 0.5 moles

If  mole of H3PO4  yield 1 mole of K3PO4

0.5 moles of H3PO4  yield 0.5 moles of K3PO4

Mass of K3PO4  = 212 g/mol * 0.5 moles = 106 g

Percent yield = 49.0 g/106 g * 100 = 46%

Q3)

Al2(SO3)3 + 6 NaOH ------> 3 Na2SO3 + 2 Al(OH)3

Number of moles of Al2(SO3)3 = 389.4 g/294 g/mol = 1.32 moles

If 1 mole of Al2(SO3)3 yields 3 moles of  Na2SO3

1.32 moles of Al2(SO3)3 yields 1.32 moles  * 3 moles/1 mole = 3.96 moles

Mass of Na2SO3 = 3.96 moles * 126 g/mol = 498.96 g

Percent yield = 212.4 g/498.96 g * 100/1 = 43%

Q4)

Al(OH)3(s) + 3 HCl(aq) -------> AlCl3(aq) + 3 H2O(l)

Number of moles of Al(OH)3 = 50.3 g/78 g/mol = 0.64 moles

If 1 mole of Al(OH)3 yields 1 mole of AlCl3

0.64 moles of Al(OH)3 yields 0.64 moles of AlCl3

Mass of AlCl3 = 133 g/mol * 0.64 moles = 85 g

Percent yield = 39.5 g/85 g * 100 = 46%

Q5) K2CO3 + 2HCl --------> H2O + CO2 + 2KCl

Number of moles of K2CO3 = 34.5 g/138 g/mol = 0.25 moles

1 mole of K2CO3  produces 2 moles of KCl

0.25 moles of K2CO3  produces 0.25 moles * 2 moles/1 mole = 0.5 moles

Mass of KCl = 0.5 moles * 75 g/mol = 37.5 g

If 1 mole of K2CO3  yields 1 mole of H2O

0.25 moles of K2CO3  yields 0.25 moles of H2O

Mass of H2O = 0.25 moles  * 18 g/mol = 4.5 g

Percent yield = 3.4 g/4.5 g *100 = 76%

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