The given mass of cobalt chloride hydrate = 2.055 g
A sample of cobalt chloride hydrate was heated to drive off waters of hydration and the anhydrate was weighed.
The mass of anhydrous cobalt chloride = 1.121 g anhydrate.
The mass of water lost during heating = 2.055 g - 1.121 g = 0.934 g
Converting mass of water of hydration present in the hydrate to moles using molar mass:
Mass of water = 0.934 g
Molar mass of water = 18.0 g/mol
Moles of water = 
Answer:
it is probably acrylic or latex
Explanation:
(133.9) x (595.296)= 79,710.1344
Answer with significant digits: 80,000
1.3 has two significant digits so 79,700 rounds up to 80000.
When The balanced equation is:
2Al + 3CuCl2 ⇒3 Cu + 2AlCl3
So, we want to find the limiting reactant:
1- no. of moles of 2Al = MV/n = (Wt * V )/ (M.Wt*n*V) = Wt / (M.Wt *n)
where M= molarity, V= volume per liter and n = number of moles in the balanced equation.
by substitute:
∴ no. of moles of 2Al = 0.2 / (26.98 * 2)= 0.003706 moles.
2- no.of moles of 3CuCl2= M*v / n = (0.5*(15/1000)) / 3= 0.0025 moles.
So, CuCl2 is determining the no.of moles of the products.
∴The no. of moles of 3Cu = 0.0025 moles.
∴The no.of moles of Cu= 3*0.0025= 0.0075 moles.
and ∵ amount of weight (g)= no.of moles * M.Wt = 0.0075 * M.wt of Cu
= 0.0075 * 63.546 =0.477 g
