Answer:
Explanation:
We are given the amounts of two reactants, so this is a limiting reactant problem.
1. Assemble all the data in one place, with molar masses above the formulas and other information below them.
Mᵣ: 58.44
NaCl + AgNO₃ ⟶ NaNO₃ + AgCl
m/g: 0.245
V/mL: 50.
c/mmol·mL⁻¹: 0.0180
2. Calculate the moles of each reactant
3. Identify the limiting reactant
Calculate the moles of AgCl we can obtain from each reactant.
From NaCl:
The molar ratio of NaCl to AgCl is 1:1.
From AgNO₃:
The molar ratio of AgNO₃ to AgCl is 1:1.
AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.
4. Calculate the moles of excess reactant
Ag⁺(aq) + Cl⁻(aq) ⟶ AgCl(s)
I/mmol: 0.900 4.192 0
C/mmol: -0.900 -0.900 +0.900
E/mmol: 0 3.292 0.900
So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.
5. Calculate the concentration of Cl⁻
![\text{[Cl$^{-}$] } = \dfrac{\text{3.292 mmol}}{\text{50. mL}} = \textbf{0.066 mol/L}\\\text{The concentration of chloride ion is $\large \boxed{\textbf{0.066 mol/L}}$}](https://tex.z-dn.net/?f=%5Ctext%7B%5BCl%24%5E%7B-%7D%24%5D%20%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B3.292%20mmol%7D%7D%7B%5Ctext%7B50.%20mL%7D%7D%20%3D%20%5Ctextbf%7B0.066%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20concentration%20of%20chloride%20ion%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.066%20mol%2FL%7D%7D%24%7D)
Answer:
4.65 L of NH₃ is required for the reaction
Explanation:
2NH₃(g) + H₂SO₄(aq) → (NH₄)₂SO₄(s)
We determine the ammonium sulfate's moles that have been formed.
8.98 g . 1mol / 132.06 g = 0.068 moles
Now, we propose this rule of three:
1 mol of ammonium sulfate can be produced by 2 moles of ammonia
Therefore, 0.068 moles of salt were produced by (0.068 . 29) / 1 = 0.136 moles of NH₃. We apply the Ideal Gases Law, to determine the volume.
Firstly we do unit's conversions:
27.6°C +273 = 300.6 K
547.9 mmHg . 1 atm / 760 mmHg = 0.721 atm
V = ( n . R . T ) / P → (0.136 mol . 0.082 L.atm/mol.K . 300.6K) / 0.721 atm
V = 4.65 L
<u>Answer:</u> The pressure of carbon dioxide gas is 11 atm
<u>Explanation:</u>
To calculate the pressure of gas, we use the equation given by ideal gas equation:
PV = nRT
where,
P = pressure of the gas = ?
V = Volume of gas = 25 L
n = number of moles of gas = 10 mole
R = Gas constant = 
T = temperature of the gas = 325 K
Putting values in above equation, we get:
Hence, the pressure of carbon dioxide gas is 11 atm
Answer:
(1) order = 2
(2) R = K [A]²
Explanation:
Given the reaction:
A--------->Product
The rate constant relation for the reaction is given as:
R(i) = K [A]............(*)
Where R(I) is rate constant at different concentration of A.
Taking the rate constant as R1, R2 and R3 for the different concentrations respectively. Then the following equations results
0.011 = K [0.15] ⁿ.........(1)
0.044 = K [0.30]ⁿ .......(2)
0.177 = K [0.60]ⁿ .........(3)
Dividing (2) by (1) and (3) by (1)
Gives:
0.044/0.011 = [0.3/0.15]ⁿ
4 = 2ⁿ; 2² = 2ⁿ; n = 2
Similarly
0.177/0.011 = [0.60/0.15]ⁿ
16.09 = 4ⁿ
16.09 = 16 (approximately)
4² = 4ⁿ ; n = 2
Hence the order of the reaction is 2.
The rate law is R = K [A]²
