How many grams of hydrogen iodide are needed when 0.687 moles of hydrogen formed?
___Mn + ___HI -> ___MnI3 + ___H₂

[tex]2mn + \: 6hi \: - > 2mni3 \: + 3h2 \\ xg \: hi \: = 0.687mol \: h2 \times \frac{6mol \: hi}{3mol \: h2} \times \frac{127 \: g \: hi}{1mol \: hi} = 174.5 \: g \: hi[/tex]