The percent yield obtained from the stoichiometry of the reaction when the moles of the reactants and products are known.
<h3>What is stoichiometry?</h3>
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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