Concept explainers
(a)
Interpretation:
Balanced equation for the reaction given has to be written.
CuSO4⋅5H2O(s)+PH3(g)→Cu3P2(s)+H2SO4(aq)+H2O(l)
(a)
Explanation of Solution
Skeletal equation is given as follows.
CuSO4⋅5H2O(s)+PH3(g)→Cu3P2(s)+H2SO4(aq)+H2O(l)
Balancing copper atoms:
In the above equation, there is one copper atom on the left side while three copper atoms are present in the right side of equation. Adding coefficient 3 before CuSO4⋅5H2O in the reactant side balances the copper atoms on both sides of the equation. The chemical equation obtained is given as shown below;
3CuSO4⋅5H2O(s)+PH3(g)→Cu3P2(s)+H2SO4(aq)+H2O(l)
Balancing sulfate ions:
In the above equation, there are three sulfate ions on the left side of equation while there is one sulfate ion on the product side of equation. Adding coefficient 3 before H2SO4 balances the sulfate ions on both sides of the equation. The chemical equation obtained is given as shown below;
3CuSO4⋅5H2O(s)+PH3(g)→Cu3P2(s)+3H2SO4(aq)+H2O(l)
Balancing phosphorus atoms:
In the above equation, there is one phosphorus atom on the left side of equation while there are two phosphorus atoms on the product side of equation. Adding coefficient 2 before PH3 balances the phosphorus atoms on both sides of the equation. The chemical equation obtained is given as shown below;
3CuSO4⋅5H2O(s)+2PH3(g)→Cu3P2(s)+3H2SO4(aq)+H2O(l)
Balancing water molecules:
In the above equation, there are fifteen water molecules on the left side of equation while there is one water molecule on the product side of equation. Adding coefficient 15 before H2O balances the water molecules on both sides of the equation. The balanced chemical equation obtained is given as shown below;
3CuSO4⋅5H2O(s)+2PH3(g)→Cu3P2(s)+3H2SO4(aq)+15H2O(l)
(b)
Interpretation:
Name of the reactants and products in the reaction has to be given.
(b)
Explanation of Solution
The balanced chemical equation is given as follows;
3CuSO4⋅5H2O(s)+2PH3(g)→Cu3P2(s)+3H2SO4(aq)+15H2O(l)
Name of the reactants:
The reactant CuSO4⋅5H2O is named as copper(II) sulfate pentahydrate and PH3 is named as phosphine.
Names of the products:
The product Cu3P2 is named as copper triphosphide, H2SO4 is named as sulfuric acid and H2O is named as water.
(c)
Interpretation:
Limiting reactant in the reaction has to be determined.
(c)
Explanation of Solution
The balanced chemical equation is given as follows;
3CuSO4⋅5H2O(s)+2PH3(g)→Cu3P2(s)+3H2SO4(aq)+15H2O(l)
Mass of CuSO4⋅5H2O taken is 0.110 kg and that of phosphine is 4.94 g that is 85.0 %. The molar mass of CuSO4⋅5H2O is 249.68 g⋅mol−1. The number of moles of CuSO4⋅5H2O and PH3 is calculated as follows;
n(CuSO4⋅5H2O)=110 g249.68 g⋅mol−1=0.44 moln(PH3)=4.94 g33.99 g⋅mol−1×85100=0.123 mol
From the balanced chemical equation, it is found that three moles of CuSO4⋅5H2O reacts with two moles of phosphine to produce one mole of Cu3P2. The amount of Cu3P2 can be calculated as shown below;
n(Cu3P2)=0.44 mol of CuSO4⋅5H2O×1 mol of Cu3P23 mol of CuSO4⋅5H2O=0.147 mol of Cu3P2n(Cu3P2)=0.123 mol of PH3×1 mol of Cu3P22 mol of PH3=0.062 mol of Cu3P2
From the above calculation it is found that phosphine is the limiting reactant while CuSO4⋅5H2O is the excess reagent.
(d)
Interpretation:
Mass of Cu3P2 produced if the percentage yield is 6.31 % has to be calculated.
(d)
Explanation of Solution
The limiting reactant is found to the PH3 and the moles of Cu3P2 that is formed is calculated to be 0.062 mol. Thus the theoretical yield of Cu3P2 formed can be calculated as follows.
Theoretical yield=0.062 mol ×252.56 g⋅mol−1=15.66 g
The percentage yield of reaction is given as 6.31 %.
Actual yield can be calculated as follows.
Actual yield=Percentage yield×Theoretical yield100=6.31×15.66 g100=0.99 g
Thus the mass of Cu3P2 produced is 0.99 g.
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Chapter F Solutions
ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM
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