Answer:
0.8 mol.
Explanation:
- The balanced equation for the reaction between Al and FeO is represented as:
<em>2Al + 3FeO → 3Fe + Al₂O₃,</em>
It is clear that 2 mol of Al react with 3 mol of FeO to produce 3 mol of Fe and 1 mol of Al₂O₃.
<em><u>Using cross multiplication:</u></em>
2 mol of Al needs → 3 mol of FeO, from stichiometry.
??? mol of Al needs → 1.2 mol of FeO.
∴<em> The no. of moles of Al are needed to react completely with 1.2 mol of FeO </em>= (2 mol)(1.2 mol)/(3 mol) = <em>0.8 mol.</em>
Answer:
<h2>Density = 0.5 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
<h3>

</h3>
From the question
mass = 60 g
volume = 120 mL
Substitute the values into the above formula and solve
That's
<h3>

</h3>
We have the final answer as
<h3>Density = 0.5 g/mL</h3>
Hope this helps you
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