Determine the mass of the salt obtained if 84 g of magnesium chloride and 86 g of lithium hydroxide are mixed.

Given:
m1 (MgCl2) – 84g.
m2 (LiOH) -86g.
To find:
m (salt) -? g.
Formulas:
m (salts) -M • n.
m1 + m2 = m (total).
M (in-va) -atomic mass • amount of substance.
M (common) -M1 + M2
n = m / M
Decision:
m (total) = m (MgCl2) + m (LiOH) = 84 + 86 = 170g.
M (MgCl2) = 24 • 1 + 35.5 • 1 = 59.5 g / mol
M (LiOH) = 7 • 1 + 16 • 1 + 1 • 1 = 24g / mol
M (total) = 59.5+ 24 = 83.5g / mol
n (salt) = m / M
n (salt) = 170 / 83.5 = 2.036 mol.
m (salt) = 83.5 g / mol • 2.036 mol = 170.006 g.