CuCl2 + 2NaNO3 ----> Cu(NO3)2 + 2NaCl
using molar masses:-
Theoretical yields:-
63.54 + 2(35.45) g of CuCl2 produces 2(22.98 + 35.45) g of NaCl
134.44 g .................................................... 116.86 g
31.0 g ....................................................31.0 * 116.86 /134.44=26.95g
So percentage yield is 21.2* 100 / 26.95 = 78.7% to nearest tenth
Answer:
177.1 L
Explanation:
The excersise can be solved, by the Ideal Gases Law.
P . V = n . R . T
In first step we need to determine the moles of gas:
We convert T° from, C° to K → 20°C + 273 = 293K
We convert P from mmHg to atm → 760 mmHg = 1atm
1Dm³ = 1L → 190L
We replace: 190 L . 1 atm = n . 0.082 . 293K
(190L.atm) / 0.082 . 293K = 7.91 moles.
We replace equation at STP conditions (1 atm and 273K)
V = (n . R .T) / P
V = (7.91 mol . 0.082 . 273K) / 1atm = 177.1 L
We can also make a rule of three:
At STP conditions 1 mol of gas occupies 22.4L
Then, 7.91 moles will be contained at (7.91 . 22.4) /1 = 177.1L
Answer is: formula is Al₂(CO₃)₃.
Aluminium carbonate (Al₂(CO₃)₃) has neutral charge. Because aluminium cation has positive charge 3+ and carbonate anion has negative charge 2-, for right chemical formula, we need two aluminium cations and three carbonate anion:
charge of the molecule = 2 · (3+) + 3 · (-2).
charge of the molecule = 0.
Since medals form cations
nonmedals form anions
