The solubility of potassium chloride in at room temperature is approximately 34 g per 100 g of water. Therefore, the maximum amount that could be dissolved would be 34/100 ( 200) = 68 g of KCl. When more than this amount is added, excess potassium would not dissolve forming crystals in the solution.
It's 18 (the same as the number of protons:)
Answer:
36.55kJ/mol
Explanation:
The heat of solution is the change in heat when the KNO3 dissolves in water:
KNO3(aq) → K+(aq) + NO3-(aq)
As the temperature decreases, the reaction is endothermic and the molar heat of solution is positive.
To solve the molar heat we need to find the moles of KNO3 dissolved and the change in heat as follows:
<em>Moles KNO3 -Molar mass: 101.1032g/mol-</em>
10.6g * (1mol/101.1032g) = 0.1048 moles KNO3
<em>Change in heat:</em>
q = m*S*ΔT
<em>Where q is heat in J,</em>
<em>m is the mass of the solution: 10.6g + 251.0g = 261.6g</em>
S is specififc heat of solution: 4.184J/g°C -Assuming is the same than pure water-
And ΔT is change in temperature: 25°C - 21.5°C = 3.5°C
q = 261.6g*4.184J/g°C*3.5°C
q = 3830.87J
<em>Molar heat of solution:</em>
3830.87J/0.1048 moles KNO3 =
36554J/mol =
<h3>36.55kJ/mol</h3>
<em />
Answer:
87.9%
Explanation:
Balanced Chemical Equation:
HCl + NaOH = NaCl + H2O
We are Given:
Mass of H2O = 9.17 g
Mass of HCl = 21.1 g
Mass of NaOH = 43.6 g
First, calculate the moles of both HCl and NaOH:
Moles of HCl: 21.1 g of HCl x 1 mole of HCl/36.46 g of HCl = 0.579 moles
Moles of NaOH: 43.6 g of NaOH x 1 mole of NaOH/40.00 g of NaOH = 1.09 moles
Here you calculate the mole of H2O from the moles of both HCl and NaOH using the balanced chemical equation:
Moles of H2O from the moles of HCl: 0.579 moles of HCl x 1 mole of H2O/1 mole of HCl = 0.579 moles
Moles of H2O from the moles of NaOH: 1.09 moles of HCl x 1 mole of H2O/1 mole of NaOH = 1.09 moles
From the calculations above, we can see that the limiting reagent is HCl because it produced the lower amount of moles of H2O. Therefore, we use 0.579 moles and NOT 1.09 moles to calculate the mass of H2O:
Mass of H2O: 0.579 moles of H2O x 18.02 g of H2O/1 mole of H2O = 10.43 g
% yield of H2O = actual yield/theoretical yield x 100= 9.17 g/10.43 g x 100 = 87.9%
Answer:
89.34%
Explanation:
First, write a balanced reaction.
Mg3N2 + <u>6</u>H2O --> <u>3</u>Mg (OH)2 + <u>2</u>NH3
Next determine the moles of the known substance, or limiting reagent ( H2O)
n= m/MM
n ( H2O) = 4.33/(1.008×2)+16
n(H2O)= 0.2403
Use the mole ratio to find the moles of Mg(OH)2
0.2403 ÷2
n (Mg (OH)2) = 0.1202
Next, find the theoretical mass of Mg (OH)2 that should have been produced
m= n × MM
m= 0.1202 × (24.305 + (16×2) +(1.008 ×2))
=7.007g
To find percentage yield, divide the experimental amount by the theoretical amount and multiply by 100.
6.26/ 7.007 × 100
=89.34%
