Answer:
18.76 g of copper II nitrate
Explanation:
Now recall that we must use the formula;
n= CV
Where;
n= number of moles of copper II nitrate solid
C= concentration of copper II nitrate solution
V= volume of copper II nitrate solution
Note that;
n= m/M
Where;
m= mass of solid copper II nitrate
M= molar mass of copper II nitrate
Thus;
m/M= CV
C= 0.05 M
V= 2.00 L
M= 187.56 g/mol
m= the unknown
Substituting values;
m/ 187.56 g/mol = 0.05 M × 2.00 L
m= 0.05 M × 2.00 L × 187.56 g/mol
m= 18.76 g of copper II nitrate
Therefore, 18.76 g of copper II nitrate is required to make 0.05 M solution of copper II nitrate in 2.00 L volume.
Answer: 233 L of 
will be produced from 150 grams of 
Explanation:
To calculate the moles :
The balanced chemical equation is:
According to stoichiometry :
2 moles of produce = 8 moles of
Thus 2.59 moles of will produce=
of
Volume of 
Thus 233 L of will be produced from 150 grams of
Answer:
0.46M NaS₂O₃ (Assuming KIO₃ solution with a concentration of 1.0M)
Explanation:
Based on the reaction:
6 Na₂S₂O₃ + KIO₃ + 5 KI + 3 H₂SO₄ → 3 Na₂S₄O₆ + 3 H₂O + 3 K₂SO₄ + 6 NaI
<em>6 moles of Na₂S₂O₃ react per mole of KIO₃</em>
Assuming the molarity of the KIO₃ solution is 0,1M:
Moles of KIO₃: = 5.0x10⁻³L ₓ (0.1 mol / L) = <em>5.0x10⁻⁴ moles</em>
As 6 moles of thiosulfate reacted per mole of iodate:
5.0x10⁻⁴ moles KIO₃ ₓ (6 moles Na₂S₂O₃ / 1 mole KIO₃) =
<em>3.0x10⁻³ moles of Na₂S₂O₃. </em>In 6.5mL (6.5x10⁻³L):
3.0x10⁻³moles Na₂S₂O₃ / 6.5x10⁻³ L = 0.46M NaS₂O₃
