Humboldt State University ® Department of Chemistry

Richard A. Paselk

	General Chemistry	Fall 2009
	© R. Paselk 2008
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Reaction Stoichiometry - Limiting Problems

1. Given the following reaction:

NH_{3 (g)} + O_2(g) NO_(g)+ H₂O_(g)

Find the maximum amount of NO in grams that can be formed from 80.0 g NH₃ and 156 g of O₂.

Our first steps are to balance the equation and to find the moles of the reactants.

Balancing by inspection:

First we see that N is balanced:

NH_{3 (g)} + O_2(g) NO_(g)+ H₂O_(g)

Next balance hydrogen. This would give 3/2 for the coeficient of water. Multiplying by 2 to get whole number coefficients we then have:

2 NH_{3 (g)} + 2 O_2(g) 2 NO_(g)+ 3 H₂O_(g)

Finally balance oxygen. This would give 5/2 oxygen. Multiplying by 2 to get whole number coefficients we then have the final equation:

4 NH_{3 (g)} + 5 O_2(g) 4 NO_(g)+ 6 H₂O_(g)

Find how many moles we have of ammonia and of oxygen.

From the Periodic table we can read the atomic weights of nitrogen and hydrogen to calculate the mole of ammonia:

IA

IIA

1 H 1.008

IIIA

IVA

VA

VIA

VIIA

VIIIA

7 N 14.01

IIIB

IVB

VB

VI

VIIB

VIIIB

IB

IIB

Moles NH₃ = 80.0 g/(14.01 + 3x1.008) g/mol = 4.697 mol

From the Periodic table we can read the atomic weight of oxygen to calculate the mole of ammonia:

IA

IIA

IIIA

IVA

VA

VIA

VIIA

VIIIA

8 O 16.00

IIIB

IVB

VB

VI

VIIB

VIIIB

IB

IIB

Moles O₂ = 156 g/(2x16.00) g/mol = 4.875 mol

Determination of limiting substance

Now we can determine which limits by calculating the moles of NO which could be made with each substance by assuming it is limiting (that is that there is excess of the other).

Assume NH₃ limits. From the stoichiometry moles NO = moles NH₃:

4 NH_{3 (g)} + 5 O_2(g) 4 NO_(g)+ 6 H₂O_(g)

So if NH₃ limits moles NO = (4 mol NO/4 mol NH₃)(4.697 mol NH₃) = 4.697 mol.

Assume O₂ limits. From the stoichiometry 4 moles NO = 5 moles O₂, or get 4 mol NO per mol of O₂:

4 NH_{3 (g)} + 5 O_2(g) 4 NO_(g)+ 6 H₂O_(g)

So if O₂ limits then moles NO = (4 mol NO/5 mol O₂)(4.875 mol NH₃) = 3.900 mol.

The smaller yield occurs when we assume O₂ limits, thus O₂ limits and,

grams NO = (3.900 mol NO)(30.01 g NO/mol NO) = 117.04 g = 117 g NO

 

2. Given the following reaction:

P_4(s) + F _2(g) PF_3(g)

Find the maximum amount of PF₃ in grams that can be formed from 122 g P₄ and 219 g of F ₂.

Our first steps are to balance the equation and to find the moles of the reactants. Balancing by inspection:

P_4(s) + 6 F _2(g) 4 PF_3(g)

Moles P₄ = 122 g/(123.88) g/mol = 0.9848 mol

Moles F ₂ = 219 g/(38.00) g/mol = 5.763 mol

Now we can determine which limits by calculating the moles of PF₃ which could be made with each substance assuming it is limiting (that is that there is excess of the other).

If P₄ limits then moles PF₃ = (4 mol PF₃/mol P₄)(0.9848 mol P₄) = 3.9392 mol.

If F ₂ limits then moles PF₃ = (4 mol PF₃/6 mol F₂)(5.763 mol F₂) = 3.842 mol.

So F₂ limits and grams PF₃ = (3.842 mol PF₃)(87.97g PF₃/mol PF₃) = 337.98 g = 338 g PF₃

 

3. Given the following reaction:

Al_(s) + Fe₂O_{3 (s)} Fe_(s) + Al ₂O_{3 (s)}

Find the maximum amount of iron in grams that can be formed from 10.6 g of aluminum and 59.4 g of iron(III) oxide.

Our first steps are to balance the equation and to find the moles of the reactants. Balancing by inspection:

2Al_(s) + Fe₂O_{3 (s)} 2Fe_(s) + Al ₂O_{3 (s)}

Moles Al = 10.6 g/(26.98) g/mol = 0.39288 mol

Moles Fe₂O₃ = 59.4 g/(159.70) g/mol = 0.37195 mol

We now have a 2:1 ratios for aluminum to iron(III) oxide and so need twice as many moles of Al as Fe₂O₃. As a result the number of moles of Al limits and we can make a maximum of 0.3988 moles of Fe since the reaction is 1:1 Fe:Al.

Grams Fe = (55.85 g Fe/mol Fe)(0.3988 moles Fe) = 22.273 g = 22.3 g

© R A Paselk

Last modified 27 October 2008