Ca as a limiting reactant
<h3>Further explanation</h3>
Given
8 g Calcium
50 g HCl
<h3>Required</h3>
Limiting reactant
Solution
Reaction
Ca + 2HCl → CaCl₂ + H₂
mol Ca (Ar = 40 g/mol) :
= mass : Ar
= 8 g : 40 g/mol
= 0.2
mol HCl (MW= 36.5 g/mol) :
= mass : MW
= 50 g : 36.5 g/mol
= 1.37
Mol : coefficient reactants :
Ca = 0.2/1 = 0.2
HCl = 1.37/2 = 0.685
Ca as a limiting reactant(smaller ratio)
Answer:
5.4 M.
Explanation:
- At complete neutralization: It is known that the no. of millimoles of acid equal that of the base.
<em>(MV)acid = (MV)NaOH</em>
M of acid = ??? M, V of acid = 35.0 mL.
M of NaOH = 3.0 M, V of NaOH = 63.0 mL.
∴ M of acid = (MV)NaOH / (V)acid = (3.0 M)(63.0 mL)/(35.0 mL) = 5.4 M.
The turbine would stop generating electricity
Answer:
At end point there will a transition from pink to colorless.
Explanation:
As the student put the vinegar in the titrator and NaOH in the beaker, it means that he has poured phenolphthalein in the NaOH solution.
The pH range of phenolphthalein is 8.3-10 (approx), it means it will show pink color in basic medium.
So on addition of phenolphthalein in NaOH the solution will become pink in color.
When we start pouring vinegar from titrator neutralization of NaOH will begin.
On complete neutralization , on addition of single drop of vinegar the solution will become acidic and there will be complete disappearance of pink color solution in the beaker.
