SO₄²⁻ +NH₃ → SO₃²⁻ + H₂O +N₂
The balanced of the above redox reaction is as below
3SO₄²⁻ + 2NH₃ → 3SO₃²⁻ + 3 H₂O + N₂
Explanation
According to the law of mass conservation the number of atoms in the reactant side must be equal to number of atoms in product side.
Inserting coefficient 3 in front of SO₄² , 2 in front of NH₃, 3 in front of SO₃²⁻ and 3 in front of H₂O balance the equation above. This is because the number of atoms are equal in both side.
for example there are 2 atoms of N in both side of the reaction.
The correct answer is C. Colligative properties only depend upon the number of solute particles in a solution but not on the identity or nature of the solute and solvent particles. I hope this anwers your question.
Answer:
30.8 grams of magnesium hydroxide will form from this reaction, and magnesium nitrate is the limiting reagent.
Explanation:
The reaction that takes place is:
- 2NaOH + Mg(NO₃)₂ → 2NaNO₃ + Mg(OH)₂
Now we <u>convert the given masses of reactants to moles</u>, using their respective <em>molar masses</em>:
- 68.3 g NaOH ÷ 40 g/mol = 1.71 mol NaOH
- 78.3 g Mg(NO₃)₂ ÷ 148.3 g/mol = 0.528 mol Mg(NO₃)₂
0.528 moles of Mg(NO₃)₂ would react completely with (0.528 * 2) 1.056 moles of NaOH. There are more than enough NaOH moles, so NaOH is the reagent in excess and <em>Mg(NO₃)₂ is the limiting reagent.</em>
Now we <u>calculate how many Mg(OH)₂ are produced</u>, using the <em>moles of the limiting reagent</em>:
- 0.528 mol Mg(NO₃)₂ *
= 0.528 mol Mg(OH)₂
Finally we convert Mg(OH)₂ moles to grams:
- 0.528 mol Mg(OH)₂ * 58.32 g/mol = 30.8 g
Answer:
1B +4St+1Y+3lc——-> BSt4Ylc3
Explanation:
I only know the answer for the first question.