From the calculation, the percentage of water in the sodium hydroxide sample is 7%.
<h3>What is neutralization?</h3>
The term neutralization has to do with the reaction between an acid and a base to yiled salt and water.
Now we have to apply the titration formula;
CAVA/CBVB = NA/NB
CA = concentration of acid
CB = concentration of base
VA = volume of acid
VB = volume of base
NA = number of moles of acid
NB = number of moles of base
The reaction equation is; HCl + NaOH ----->NaCl + H2O
CAVANB = CBVBNA
CB = CAVANB /VBNA
CB = 34.9 * 0.2 M * 1/10 * 1
CB = 0.698 M
Number of moles = Conncentration * volume
= 0.698 M * 100/1000 L = 0.0698 moles
Mass = Number of moles * molar mass
Mass = 0.0698 moles * 40 g/mol = 2.79 g
percent of NaOH = 2.79 g/ 3g * 100/1 = 93%
Percent of water = 100- 93 = 7%
Learn more about neutralization: brainly.com/question/15395418
Answer:
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The balanced chemical reaction would be
<span>fecl2 + 2naoh = fe(oh)2(s) + 2nacl
Initial amounts of the reactants are given, so, we need to determine which of the reactants is the limiting reactant and use this amount to determine what is asked. However, what is being asked is how many of the FeCl2 is used in the reaction, showing that it is NaOH that is the limiting reactants. Thus, we just use the initial amount of NaOH and relate the substances by the chemical reaction as follows:
6 mol NaOH ( 1 mol FeCl2 / 2 mol NaOH ) = 3 mol FeCl2
Therefore, 3 moles of FeCl2 is used up and 3 moles of FeCl2 is also left after the reaction.</span>
The question incomplete , the complete question is:
A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.
Answer:
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Explanation:
Moles of urea = 
Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)
Molarity of the urea solution ;
Mass of solvent = m
Volume of solvent = V = 200.0 mL
Density of the urea = d = 0.95 g/mL
(1 g = 0.001 kg)
Molality of the urea solution ;
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Answer:
1. 4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2
2. 6 moles of Cl2
Explanation:
1. The balanced equation for the reaction. This is illustrated below:
4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2
2. Determination of the number of mole of Cl2 produce when 4 moles of FeCl3 react with 4 moles. To obtain the number of mole of Cl2 produced, we must determine which reactant is the limiting reactant.
This is illustrated below:
From the balanced equation above,
4 moles of FeCl3 reacted with 3 moles of O2.
Since lesser amount of O2 (i.e 3 moles) than what was given (i.e 4 moles) is needed to react completely with 4 moles of FeCl3, therefore FeCl3 is the limiting reactant and O2 is the excess reactant.
Finally, we can obtain the number of mole Cl2 produced from the reaction as follow:
Note: the limiting reactant is used as it will produce the maximum yield of the reaction since all of it is used up in the reaction.
From the balanced equation above,
4 moles of FeCl3 will react to produced 6 moles of Cl2.
