Question: In the Reaction: CuCl2 + 2 NaNO3 -- Cu(NO3)2 + 2 Naci
If 17.2 grams of copper (II) chloride, CuCl2, reacts with 22.6 grams of sodium nitrate, NaNO3. how many grams of sodium chloride, NaCl, will be
produced? (4 points)
Molar mass of CuCl2 = 134.45 g/mol
Molar mass of NaNO3 = 85.00 g/mol
Molar mass of NaCl = 58.44 g/mol
Answers:
17.2 grams CuCl2 =
gNaci
v
22.6 grams NaNO3 =
g Naci
What is the Limiting Reactant?
No

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.