<span>13.7 g
The balanced formula is:
Pb(ClO3)2(aq)+2NaI(aq) ==> PbI2(s) + 2NaClO3(aq)
The number of moles of NaI we have is the volume of fluid multiplied by the molarity so
0.350 * 0.170 = 0.0595 moles
Since the NaI is the limiting reactant, for every two moles used, we'll produce 1 mole of precipitate. So
0.0595 mole / 2 = 0.02975 mole
Now we need to calculate the molar mass of PbI2. Looking up the atomic weights
Atomic weight Lead = 207.2
Atomic weight iodine = 126.90447
Molar mass PbI2 = 207.2 + 2 * 126.90447 = 461.00894 g/mol
Now multiply the molar mass by the number of moles we have.
461.00894 g/mol * 0.02975 mol = 13.71501597 g
Rounded to 3 significant figures, the answer is 13.7 g</span>
The correct answer should be Letter B
<u>Answer:</u> The products of the given chemical equation are
<u>Explanation:</u>
Protonation equation is defined as the equation in which protons get added in the substance.
The chemical equation for the protonation of carbonate ion in the presence of water follows:
By Stoichiometry of the reaction:
1 mole of carbonate ion reacts with 1 mole of water to produce 1 mole of hydrogen carbonate ion and 1 mole of hydroxide ion
Hence, the products of the given chemical equation are
The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2M was neutralized by 0.01 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)
<u>Answer:</u> The volume of HCl neutralized is 0.125 mL
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is HCl (Stomach acid)
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Hence, the volume of HCl neutralized is 0.125 mL