Answer:
14g ZnCl₂
Explanation:
We can convert grams of reactant to grams of product by converting to moles, and using the stoichiometric ratios in the reaction equation:
\(\dfrac{6.5g\text{ Zn}}{1} \times \dfrac{1\text{ mole Zn}}{65.38g\text{ Zn}}\times \dfrac{1\text{ mole ZnCl}_2}{1\text{ mole Zn}} \times \dfrac{(65.38+2(35.45))g\text{ ZnCl}_2}{1\text{ mole ZnCl}_2}\)
\(\approx 13.549g\text{ ZnCl}_2\)
\(\approx 14g\text{ ZnCl}_2\) ... rounded to 2 significant figures
