<span>Answer:
It depends on what came after "0.5440 M H...".
If it was a monoprotic acid, like HCl, the calculation would go like this:
(55.25 mL) x (0.5440 M acid) x (1 mol KOH / 1 mol acid) / (0.2450 M KOH) =
122.7 mL KOH
If it was a diprotic acid, like H2SO4, like this:
(55.25 mL) x (0.5440 M acid) x (2 mol KOH / 1 mol acid) / (0.2450 M KOH) =
245.4 mL KOH
If it was a triprotic acid, like H3PO4, like this:
(55.25 mL) x (0.5440 M acid) x (3 mol KOH / 1 mol acid) / (0.2450 M KOH) =
368.0 mL KOH</span>
Molarity is one of the method of expressing concentration of solution. Mathematically it is expressed as,
Molarity =
Given: Molarity of solution = 5.00 M
Volume of solution = 750 ml = 0.750 l
∴ 5 =
∴
number of moles = 3.75Answer: Number of moles of KOH present in solution is 3.75.
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
