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:
%age Yield = 85.36 %
Solution:
The Balance Chemical Reaction is as follow,
C₆H₁₂O + Acid Catalyst → C₆H₁₀ + Acid Catalyst + H₂O
According to Equation ,
100 g (1 mole) C₆H₁₂O produces = 82 g (1 moles) of C₆H₁₀
So,
4.0 g of C₆H₁₂O will produce = X g of C₆H₁₀
Solving for X,
X = (4.0 g × 82 g) ÷ 100 g
X = 3.28 g of C₆H₁₀ (Theoretical Yield)
As we know,
%age Yield = (Actual Yield ÷ Theoretical Yield) × 100
%age Yield = (2.8 g ÷ 3.28 g) × 100
%age Yield = 85.36 %