Answer:
[tex]\large \boxed{\text{129 g}}[/tex]
Step-by-step explanation:
We can use ratio and proportion to solve this problem.
Let x = the mass of Zn
[tex]\begin{array}{cccl}\dfrac{\text{Zn}}{\text{Cu}} & = & \dfrac{3}{11} & \\\\\dfrac{x}{473} & = & \dfrac{3}{11} & \text{Substituted the mass of Cu}\\\\x & = & \dfrac{3\times473}{11} &\text{Multiplied each side by 473} \\\\ & = &\mathbf{129} & \text{Simplified}\\\end{array}\\\text{The mass of Zn is $\large \boxed{\textbf{129 g}}$}[/tex]
Check:
[tex]\begin{array}{rcl}\dfrac{129}{473} & = &\dfrac{3}{11} \\\\\dfrac{3}{11} & = & \dfrac{3}{11} \\\end{array}[/tex]
OK
