To calculate the grams of \(ZnCl_2\) formed, we need to determine the limiting reactant. This can be done by comparing the moles of each reactant based on the given masses and their respective molar masses.

1. Calculate the moles of \(Zn\):

\(Moles\ of\ Zn = 150\ g/65.38\ g/mol = 2.29 mol\)

2. Calculate the moles of \(HCl\):

\(Moles\ of\ HCl = 73\ g/36.46\ g/mol = 2.00 mol\)

3. Determine the limiting reactant:

Comparing the moles of \(Zn\) and \(HCl\), we can see that \(HCl\) is the limiting reactant because it has fewer moles available. This means that all of the \(HCl\) will react, and some \(Zn\) will be left unreacted.

4. Calculate the theoretical yield of \(ZnCl_2\):

From the balanced chemical equation \(Zn + 2HCl → ZnCl_2 + H_2\), we can see that 1 mole of \(Zn\) reacts with 2 moles of \(HCl\) to produce 1 mole of \(ZnCl_2\). Therefore, 2.00 mol of \(HCl\) will produce 1.00 mol of \(ZnCl_2\).

\(Moles\ of\ ZnCl_2 = 2.00\ mol\ HCl * (1\ mol\ ZnCl_2 / 2\ mol\ HCl) = 1.00\ mol\)

5. Calculate the mass of \(ZnCl_2\) formed:

\(Mass\ of\ ZnCl_2 = 1.00\ mol * 136.29\ g/mol = 136.29\ g\)

Therefore, the theoretical yield of \(ZnCl_2\) formed from 150 g \(Zn\) and 73 g \(HCl\) is 136.29 g.