The diluted lead (II) nitrate solution from part b is mixed with 50.0 mL of 0.650 mol L−1 sodium iodide. The products of this reaction are a bright yellow precipitate of lead (II) iodide and a solution of sodium nitrate. Determine the amount, in mol, of lead (II) nitrate and sodium iodide that were mixed together. (2 marks)
Determine whether lead (II) nitrate or sodium iodide was limiting in this reaction.
(2 marks)
Determine the mass of precipitate that forms. (2 marks)
please help

The answer to that would be magnesium

For cyclohexane compounds drawn in a chair conformation:   -axial bonds are vertical  -equatorial bonds are parallel to two bonds within the ring.  thus, as i assume the chair conformation is drawn:   -the x group is in an axial position, as are the five other vertical bonds.  -three axial bonds are on the top face of the ring  -the other three axial bonds are on the bottom face  the position of the second substituent on each ring carbon is equatorial.  gauche groups must be on an adjacent carbon such as carbon 6 and carbon 2, and also gauche groups are rotated by 60 degrees relative to each other.  drawing the newman projections for c1–c2 and c1–c6 aids in visualizing the groups that are gauche to the y group on c1.  click this link and watch some academic videos that may you more understand the topic