Answer:
Option E is correct. none of the above is correct
Explanation:
Step 1: Data given
Solid Iron (III) = Fe^3+
iron (III) oxide = Fe2O3
Molar mass Fe = 55.845 g/mol
Molar mass Fe2O3 = 159.69 g/mol
Step 2: The balanced equation:
4Fe + 3O2 → 2Fe2O3
4 moles of iron will need 2 moles of oxygen gas to fully react
⇒ This is false 4 moles of iron will need 3 moles of oxygen gas to fully react
B.12 moles of iron, if reacted completely, can produce 8 moles of iron (III) oxide.
⇒ This is false: When 12 moles of iron completely react, we can produce 12/2 = 6 moles of Fe2O3
C.9 moles of oxygen can produce 9 moles of Iron (III) oxide
⇒ This is false; 9 moles of O2 can produce 6 moles of Fe2O3
D.6 moles of oxygen can react completely to produce 6 moles of iron (III) oxide.
⇒ This is false 6 moles of O2 will react completely to produce 4 moles of Fe2O3
E.none of the above
Following chemical reaction is involved upon titration of Ca(OH)2 with HCl,
Ca(OH)2 + 2HCl ↔ CaCL2 + 2H2O
Above is an example of acid-base titration to generate salt and water. Here, H+ ions of acid (HCl) combines with OH- (ions) of base [Ca(OH)2] to generated H2O
Given,
concentration of HCl = 0.0199 M
Total volume of HCl consumed during titration = 16.08 mL = 16.08 X 10^(-3) L
∴, number of moles of H+ consumed = Molarity X Vol. of HCl (in L)
= 0.0199 X 16.08 X 10^(-3)
= 3.1999 X 10^-4 mol
Thus, total number of moles of [OH-] ions present initial = 3.1999 X 10-4 mol
So, initial conc. [OH-] ion = ![\frac{number of moles of [OH-]}{volume of solution (L)}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bnumber%20of%20moles%20of%20%5BOH-%5D%7D%7Bvolume%20of%20solution%20%28L%29%7D%20)
=

= 0.03199 M