This problem is describe the mole-ratio composition of an allow composed by tin, lead and cadmium. Ratios are given as Sn:Pb 2.73:1.00 and Pb:Cd is 1.78:1.00, and we are asked to calculate the mass percent compositon of Pb in the allow.
In this case, according to the given information, it turns out possible realize that the following number of moles are present in the alloy, according to the aforementioned ratios:
[tex]2.73mol Sn\\\\1.00molPb\\\\\frac{1.00molPb*1.00molCd}{1.78molPb}= 0.562molCd[/tex]
Next, we calculate the masses by using each metal's atomic mass:
[tex]m_{Sn}=2.73mol*\frac{118.7g}{1mol}=324.05g\\\\ m_{Pb}=1.00mol*\frac{207.2g}{1mol}=207.2g\\\\m_{Cd}=0.562mol*\frac{112.4g}{1mol}=63.2g[/tex]
Thus, the mass percent composition of each metal is shown below:
[tex]\%Sn=\frac{324.05g}{324.05g+207.2g+63.2g} *100\%=54.5\%\\\\\%Pb=\frac{207.2g}{324.05g+207.2g+63.2g} *100\%=34.9\%\\\\\%Cd=\frac{63.2}{324.05g+207.2g+63.2g} *100\%=10.6\%[/tex]
So that of lead is 34.9 %.
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