If 12.6 mol of iron(ill) oxide and 15.2 mol of carbon react, what is the maximum

Question

If 12.6 mol of iron(ill) oxide and 15.2 mol of carbon react, what is the maximum number of moles of iton that can be recovered from the reaction? 2Fe2O3(s) + 3 as) 4 Fe (s) +3CO2(g) 25.2 QUESTION 2 What is the maximum number of grams of NO (30.01 g/mol) that can be formed from the reaciton of 18.6 g of NH3 (17.03 g/mol) with 29.4 g of 02 (32.00 g/mol)? 4NH3(g) + 502(g) 4 NO(g) + 6 H2O(l) 0.7 QUESTION3 If 10.2 mol of iron(l) oxide and 22.9 mol of carbon react, how many moles of the excess reactant will remain after the reaction? 2Fe203(s) + 3 cis) 4 Fe (s) + 3CO2(g)

Explanation / Answer

Answer – Q 1) We are given , moles of Fe2O3 = 12.6 moles , moles of C = 15.2 moles

First we need to calculate limiting reactant –

We need to calculate moles of Fe

Moles of Fe from Fe2O3

From balanced reaction –

2 moles of Fe2O3 = 4 moles of Fe

So 12.6 moles of Fe2O3 = ? moles of Fe

= 25.2 moles of Fe

Moles of Fe from C

From balanced reaction –

3 moles of C = 4 moles of Fe

So 15.2 moles of C = ? moles of Fe

= 20.3 moles of Fe

So limiting reactant is C , so maximum number of moles of Fe formed is 20.3 moles.

Q 2) We are given, mass of NH3 = 18.6 g , mass of O2 = 29.4 g

First we need to calculate moles of each reactant –

Moles of NH3 = 18.6 g / 17.03 g.mol-1

                       = 1.09 moles

Moles of O2 = 29.4 g / 32.00 g.mol-1

                       = 0.919 moles

Now we need to calculate limiting reactant –

We need to calculate moles of NO

Moles of NO from NH3

From balanced reaction –

4 moles of NH3 = 4 moles of NO

So 1.09 moles of NH3 = ? moles of NO

= 1.09 moles of NO

Moles of NO from O2

From balanced reaction –

5 moles of O2 = 4 moles of NO

So 0.919 moles of O2 = ? moles of NO

= 0.735 moles of NO

So, moles of NO lowest from O2, so limiting reactant is O2.

Moles of NO = 0.735 moles

Mass of NO = 0.735 moles x 30.01 g/mol

                    = 22.1 g

So, maximum number of grams of NO is 22.1 g.

Q 3) We are given , moles of Fe2O3 = 10.2 moles , moles of C = 22.9 moles

First we need to calculate limiting reactant –

We need to calculate moles of Fe

Moles of Fe from Fe2O3

From balanced reaction –

2 moles of Fe2O3 = 4 moles of Fe

So 10.2 moles of Fe2O3 = ? moles of Fe

= 20.4 moles of Fe

Moles of Fe from C

From balanced reaction –

3 moles of C = 4 moles of Fe

So 22.9 moles of C = ? moles of Fe

= 30.53 moles of Fe

So limiting reactant is Fe2O3 , so maximum number of moles of Fe formed is 20.4 moles.

Used moles of C –

2 moles of Fe2O3 = 3 moles of C

So 10.2 moles of Fe2O3 = ? moles of C

= 15.3 moles of C

Excess of C = 22.9 moles – 15.3 moles

                   = 7.6 moles

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.