Cunife 1 The Special Alloy

Let $x$ , $y$ and $z$ be the mass of Cu, Ni and Fe, respectively.

Sum of the masses: We know that the in both experiments the sample mass is 10.00 g. The equation is as follows: $x+y+z=10$

First experiment: By activity series, the unoxidized metal is Cu. There should be no “x” term in the equation. The stoichiometric ratio is 1:1 or moles of metal = moles of hydrogen gas. Thus the equation is as follows:

$\frac { y }{ 56.89\quad \frac { g }{ mol } } +\frac { z }{ 55.85\quad \frac { g }{ mol } } =\frac { 0.85494\quad L\times 2\quad atm }{ 0.08206\quad \frac { L\quad atm }{ mol\quad K } \times 298.15\quad K } =0.06989\quad mol$

Second experiment: For a calorimeter, ${ -q }_{ water }={ q }_{ metal }=\left( \sum _{ i }^{ n }{ { m }_{ i }{ c }_{ i } } \right) \Delta T$

${-q }_{ water }={ \left( 50.00\quad g \right) \left( 4.184\quad \frac { J }{ g\quad K } \right) \left( 11.738\quad ℃-10.00\quad ℃ \right) }=363.5896\quad J$ $363.5896\quad J={ \left[ \left( 0.39\quad \frac { J }{ g\quad K } \right) x+\left( 0.44\quad \frac { J }{ g\quad K } \right) y+\left( 0.45\quad \frac { J }{ g\quad K } \right) z \right] \left( 100.00\quad ℃-11.738\quad ℃ \right) }$

Simplifying, we get $0.39x+0.44y+0.45z=4.1194$

Thus we have a system of three equations. Solving the system, we get $x = 5.996 g$ , $y = 2.084 g$ , and $z = 1.920 g$ .

Thus the composition of Cunife 1 is 59.96% Cu, 20.84% Ni and 19.20% Fe.