The balanced chemical equation for the above reaction is as follows ;
Mg + 2HCl —> MgCl2 + H2
The stoichiometry of Mg to HCl is 1:2
This means that 1 mol of Mg reacts with 2 mol of HCl
Equal amounts of both Mg and HCl have been added. One reagent is the limiting reactant and other reactant is in excess.
Limiting reactant is the reagent that is fully used up in the reaction and the amount of Product formed depends on the amount of limiting reactant present.
In this reaction if Mg is the limiting reactant, 4.40 moles of Mg should react with 4.40x2 -8.80 moles of HCl.
But only 4.40 moles of HCl present therefore HCl is the limiting reactant that reacts with 4.40/2 = 2.20 moles of Mg
Stoichiometry of HCl to MgCl2 is 2:1
Since HCl moles reacted -4.40 mol
Then MgCl2 moles formed are 4.40/2 = 2.20 mol of MgCl2
Answer:
10
Explanation:
pH is defined as the negative logarithm of the concentration of hydrogen ions.
Thus,
pH = - log [H⁺]
Thus, from the formula, more the concentration of the hydrogen ions or more the acidic the solution is, the less is the pH value of the solution.
Thus, solution with pH = 3 will be more acidic than solution with pH =4
Thus, concentration of the [H⁺] when pH =3
3 = - log [H⁺]
[H⁺] = 10⁻³ M
For pH = 4, [H⁺] = 10⁻⁴ M
<u>hence, pH = 3 is 10 times more acidic than pH = 4</u>
Answer:
Cl₂O₇
Explanation:
For the reaction:
ClₓOₙ + H₂ → HCl + H₂O
Moles of HCl and moles of H₂O are:
HCl: 0.233g HCl ₓ (1mol / 36.46g) = 6.39x10⁻³ mol HCl
H₂O: 0.403g H₂O ₓ (1mol / 18.02g) = 2.236x10⁻² mol H₂O
As you can see, moles of HCl are equivalent to moles of Cl in the compound and moles of H₂O are equivalent to moles of O in the compound, that means:
6.39x10⁻³ mol Cl
2.236x10⁻² mol O
Empirical formula is the simplest ratio of atoms presents in a molecule. If Cl is <em>1</em>, Oxygen will be:
2.236x10⁻² mol / 6.39x10⁻³ = <em>3.5</em>
As empirical formula must be given in natural numbers, the empirical formula is:
<em>Cl₂O₇</em>
<em></em>
The correct answer is that 1.125 mol of NaOH is available, and 60.75 g of FeCl₃ can be consumed.
The mass of NaOH is 45 g
The molar mass of NaOH = 40 g/mol
The moles of NaOH = mass / molar mass
= 45 / 40
= 1.125
Thus, 1.125 mol NaOH is available
3 NaOH + FeCl₃ ⇒ Fe (OH)₃ + 3NaCl
3 mol of NaOH react with 1 mol of FeCl₃
1.125 moles of NaOH will react with x moles of FeCl₃
x = 1.125 / 3
x = 0.375 mol
0.375 mol FeCl₃ can take part in reaction
The molar mass of FeCl₃ is 162 g/mol
The mass of FeCl₃ = moles × mass
= 0.375 × 162
= 60.75 g
Thus, the amount of FeCl₃, which can be consumed is 60.75 g
