Answer to Question #163342 in General Chemistry for mike

Question

A 2.50 g sample of bronze, an alloy of copper and tin, was dissolved in sulfuric acid. The copper in the alloy reacts with sulfuric acid as shown by the following balanced reaction: 

Cu + 2H2SO4 ---> Cu2+ + SO2 + 2H2O + SO42¯

Adding KI produces CuI and triodide, I3¯ 

2Cu2+ + 5I¯ ---> 2CuI + I3¯

Finally, titrating the I3¯ with S2O32¯ 

I3¯ + 2S2O32¯ ---> 3I¯ + S4O62¯

provides an indirect method for determining the amount of Cu in the original sample. Calculate the percentage, by mass, of copper in a sample of bronze if 31.50 mL of 1.000 M S2O32¯ is consumed in the titration.                                                    

Answer 1

Combining all the above given reaction,

The overall reaction is-

$2Cu(s)+2I^-+2S_2O_3^{2-}\rightarrow 2CuI(s)+2SO_2+S_4O_6^{2-}+4H_2O+2SO_4^{2-}$

so from the above equatiom

2 moles of $cu$ = 2 moles of $S_2O_3^{2-}$

the moles of $S_2O_3^{2-}$ in $31.5\text{ml}=\dfrac{1.00\times 31.5}{1000}=0.0315 \text{ml}$

Let $x$ mole of copper react,

$x$ mole of copper=$0.0315$ mole of $S_2O_3^{2-}$

So mole of $Cu(s)$ =$0.0315 \text{mol}$

Mass of Copper=$\text{mole}\times \text{Atomic mass}$

=$0.0315\times 63.55$

=$2\text{g}$

Mass of sample=$2.5\text{g}$

Mass percent of Copper=$\dfrac{2.00}{2.50}\times 100=60$ %