Answer:
0.26g of NaCl is the maximum mass that could be produced
Explanation:
Based on the reaction:
HCl + NaOH → NaCl + H₂O
<em>Where 1 mol of HCl reacts per mol of NaOH to produce 1 mol of NaCl</em>
<em />
To solve this question we need to find <em>limiting reactant. </em>The moles of limiting reactant = Moles of NaCl produced:
<em>Moles HCl -Molar mass: 36.46g/mol-:</em>
0.365g HCl * (1mol / 36.46g) = 0.010 moles HCl
<em>Moles NaOH -Molar mass: 40g/mol-:</em>
0.18g NaOH * (1mol / 40g) = 0.0045 moles NaOH
As the reaction is 1:1 and moles NaOH < moles HCl, limiting reactant is NaOH and maximum moles produced of NaCl are 0.0045 moles.
The mass of NaCl is:
<em>Mass NaCl -Molar mass: 58.44g/mol-:</em>
0.0045 moles * (58.44g/mol) =
<h3>0.26g of NaCl is the maximum mass that could be produced</h3>
Answer:
A = 2A + 3B → 5C
Explanation:
The two molecule of A and three molecules of B will react to form the five molecules of C.
2A + 3B → 5C
Other options are incorrect because,
B = A₂ + B₃ → C₅
in this reaction one molecule of A₂ and one molecule of B₃ combine to form one molecule of C₅.
C = 2A + 5B → 3C
in this reaction two molecules of A and five molecules of B combine to form three molecule of C.
D = A₂ + B₃ → C₃
in this reaction one molecule of A₂ and one molecule of B₃ combine to from one molecule of C₃.
Answer:
The empirical formula is Ag2O.
The empirical formula is Ag2O.Explanation:
The empirical formula is Ag2O.Explanation:The empirical formula is the simplest whole-number ratio of atoms in a compound.
The empirical formula is Ag2O.Explanation:The empirical formula is the simplest whole-number ratio of atoms in a compound.The ratio of atoms is the same as the ratio of moles. So our job is to calculate the molar ratio of Ag to 2O.
do the steps ...
To get this into an integer ratio, we divide both numbers by the smaller value.
From this point on, I like to summarize the calculations in a table.
ElementAgMass/gXMolesXllRatiomllIntegers
—————————————————−———mAgXXXm7.96Xm0.07377Xll2.00mmm2
mlOXXXXl0.59mm0.0369Xml1mmmml1
There are 2 mol of Ag for 1 mol of O.
Answer:
pH = 12.33
Explanation:
Lets call HA = butanoic acid and A⁻ butanoic acid and its conjugate base butanoate respectively.
The titration reaction is
HA + KOH ---------------------------- A⁻ + H₂O + K⁺
number of moles of HA : 118.3 ml/1000ml/L x 0.3500 mol/L = 0.041 mol HA
number of moles of OH : 115.4 mL/1000ml/L x 0.400 mol/L = 0.046 mol A⁻
therefore the weak acid will be completely consumed and what we have is the unreacted strong base KOH which will drive the pH of the solution since the contribution of the conjugate base is negligible.
n unreacted KOH = 0.046 - 0.041 = 0.005 mol KOH
pOH = - log (KOH)
M KOH = 0.005 mol / (0.118.3 +0.1154)L = 0.0021 M
pOH = - log (0.0021) = 1.66
pH = 14 - 1.96 = 12.33
Note: It is a mistake to ask for the pH of the <u>acid solutio</u>n since as the above calculation shows we have a basic solution the moment all the acid has been consumed.
