Answer:
Equilibrium constant Kc for the reaction will be 1.722
Explanation:
O2(g)+NO(g)→CO(g)+ NO2(g)
0.88 3.9 --- ---
0.88x 3.9-x x x
GIVEN:
0.88X-X= 0.11
⇒ X=0.77
CO2(g)+NO(g) → CO(g) + NO2(g)
0.88 3.9 --- ---
0.88-x 3.9-x x x
= 3.13 0.77 0.77
=0.11
Kc = ![\frac{[CO] *[NO2]} {[CO2]*[NO]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCO%5D%20%2A%5BNO2%5D%7D%20%7B%5BCO2%5D%2A%5BNO%5D%7D%20)
=
= 1.722
Answer:
mole fraction of NaCl = 0.03145.
mole fraction of water = 0.9686.
Explanation:
- Mole fraction is an expression of the concentration of a solution or mixture.
- It is equal to the moles of one component divided by the total moles in the solution or mixture.
- The summation of mole fraction of all mixture components = 1.
mole fraction of NaCl = (no. of moles of NaCl) / (total no. of moles).
<em>no. of moles of NaCl = mass/molar mass </em>= (6.87 g)/(58.44 g/mol) = 0.1176 mol.
<em>no. of moles of water = mass/molar mass</em> = (65.2 g)/(18.0 g/mol) = <em>3.622 mol.</em>
<em></em>
∴ mole fraction of NaCl = (no. of moles of NaCl) / (total no. of moles) = (0.1176 mol)/(0.1176 mol + 3.622 mol) = 0.03145.
<em>∵ mole fraction of NaCl + mole fraction of water = 1.0.</em>
∴ mole fraction of water = 1.0 - mole fraction of NaCl = 1.0 - 0.03145 = 0.9686.
Answer:
%age Yield = 34.21 %
Explanation:
The balance chemical equation for the decomposition of KClO₃ is as follow;
3 KOH + H₃PO₄ → K₃PO₄ + 3 H₂O
Step 1: Calculate moles of H₃PO₄ as;
Moles = Mass / M/Mass
Moles = 334.6 g / 97.99 g/mol
Moles = 3.414 moles
Step 2: Find moles of K₃PO₄ as;
According to equation,
1 moles of H₃PO₄ produces = 1 moles of K₃PO₄
So,
3.414 moles of H₃PO₄ will produce = X moles of K₃PO₄
Solving for X,
X = 1 mol × 3.414 mol / 1 mol
X = 3.414 mol of K₃PO₄
Step 3: Calculate Theoretical yield of K₃PO₄ as,
Mass = Moles × M.Mass
Mass = 3.414 mol × 212.26 g/mol
Mass = 724.79 g of K₃PO₄
Also,
%age Yield = Actual Yield / Theoretical Yield × 100
%age Yield = 248 g / 724.79 × 100
%age Yield = 34.21 %
Answer:
Mass = 0.37 g
Explanation:
Given data:
Number of moles of sulfur = 11.9 mol
Mass of sulfur in 11.9 mol = ?
Molar mass of sulfur = 32.06 g
Solution:
Number of moles = mass/molar mass
by putting values,
11.9 mol = mass/ 32.06 g/mol
Mass = 11.9 mol × 32.06 g/mol
Mass = 0.37 g
