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1. STOICHIOMETRY INVOLVING ONLY PURE SUBSTANCES
For all chemical reactions, the balanced chemical equation gives the mole ratios of reactants and products. If we are dealing with pure chemicals, the molar mass allows us to convert the mass of a reactant or product into moles. Consider the reaction shown below.
Ca3N2( s ) +^ 6H 2 O( l ) H 2NH 3 ( g ) + 3Ca(OH) 2 ( s )
mole ratio 1 :^6 :^2 :^3 molar mass 148.3 18.0 17.0 (^) 74. (g/mole)
For this reaction, 1 mole of Ca 3 N2 will react with 6 moles of H 2 O to produce 2 moles of NH 3 and 3 moles of Ca(OH)2. Therefore, for this reaction 1 mole Ca3N2 ≡ 6 moles H2O. Similar equivalences will apply to other pairs of reactants and/or products. For example, 6 moles H 2 O ≡ 2 moles NH 3. As discussed before, each equivalence will give two conversion factors.
(a) Calculate the number of moles of H 2 O that will react with 2.5 moles of Ca 3 N 2.
Solution : moles H 2 O = = 15 moles H 2 O
6 moles H 2 O 1 mole Ca 3 N
2.5 moles Ca3N2 x
given mole ratio
(b) How many moles of NH 3 can be made by the reaction of 0.75 mole of Ca 3 N2 with an excess of H2O? ( Ans. 1.5 mole)
(c) How many moles of H 2 O are needed to make 12 moles of NH 3? ( Ans. 36 moles)
(d) Calculate the mass of Ca(OH) 2 that can be formed from the reaction of 10.0 g of Ca 3 N2 with an excess of H 2 O.
Solution: moles Ca 3 N2 = 10.0 g Ca 3 N2 x = 0.06743 mole
1 mole Ca 3 N
148.3 g Ca3N (^2)
moles Ca(OH) 2 = 0.06743 mole Ca 3 N2 x = 0.2023 mole 1 mole Ca 3 N
3 moles Ca(OH) (^2)
mass Ca(OH) 2 = 0.2023 mole x = 15.0 g
74.1 g Ca(OH) (^2)
1 mole Ca(OH) (^2)
(e) What mass of NH 3 can be formed from the reaction of 23.5 g of H 2 O with an excess of Ca 3 N2? ( Ans. 7.40 g)
(f) What mass of H 2 O is needed to make 454 g of Ca(OH) 2? ( Ans. 221 g)
(g) What volume of H 2 O is needed to react completely with 45.0 g of Ca 3 N2? ( Ans. 32.8 mL)
2. THEORETICAL, ACTUAL AND PERCENT YIELD
The amount of a product calculated for a reaction is only a theoretical amount and is therefore called the theoretical yield. When a reaction is actually performed, the amount of product obtained ( or isolated) (the actual yield) is usually less than the theoretical yield. The percent yield gives the actual yield as a percentage of the theoretical yield.
percent yield =
actual yield theoretical yield x 100
(h) If only 6.85 g of NH 3 was obtained from the reaction in (e), what is the percent yield of the reaction? ( Ans. 92.6%)
(i) If the percent yield of Ca(OH) 2 from Ca 3 N2 is 93.0%, what mass of Ca 3 N 2 must be used to obtain 66.0 g of Ca(OH) 2? (HINT: Use the percent yield to calculate the theoretical yield and then use the theoretical to calculate the amount of reactant needed). ( Ans. 47.3 g)
3. STOICHIOMETRY AND PERCENT PURITY
Many samples of chemicals are not pure. We can define percent purity as
mass of pure compound in the impure sample total mass of impure sample x 100
If an impure sample of a chemical of known percent purity is used in a chemical reaction, the percent purity has to be used in stoichiometric calculations. Conversely, the percent purity of an impure sample of a chemical of unknown percent purity can be determined by reaction with a pure compound as in an acid-base titration. Percent purity can also be determined, in theory, by measuring the amount of product obtained from a reaction. This latter approach, however, assumes a 100% yield of the product.
Consider the reaction of magnesium hydroxide with phosphoric acid.
3Mg(OH) 2 + 2H3PO 4 H Mg3(PO4)2 + 6H2O
(a) Calculate the mass of Mg 3 (PO4) 2 that will be formed (assuming a 100% yield) from the reaction of 15.0 g of 92.5% Mg(OH) 2 with an excess of H 3 PO 4.
mass Mg(OH) 2 = 15.0 x 0.925 = 13.875 g mass Mg3(PO4) 2 =
13.875 g Mg(OH) 2 x
1 mole Mg 3 (PO 4 ) x 58.3 g Mg(OH) (^2)
1 mole Mg(OH) 3 moles Mg(OH) (^2) 1 mole Mg3(PO 4 ) 2
262.9 g Mg3(PO4) x
= 20.9 g Mg3(PO4) 2
4. LIMITING REACTANTS
(a) Consider the reaction Ca 3 N 2 + 6H 2 O H 2NH 3 + 3Ca(OH) 2
For the reaction of 20.0 g of Ca 3 N 2 with 16.0 g of H2O, ( 1) calculate the theoretical yield of Ca(OH)2, (2) determine the limiting reactant, and (3) calculate the mass of the excess reactant left over.
Solution :
From Ca 3 N2: mass of Ca(OH)2 = 20.0 g Ca 3 N2 x = 30.0 g
3 x 74.1 g Ca(OH) (^2) 148.3 g Ca3N
From H2O: mass of Ca(OH)2 = 16.0 g H 2 O x = 32.9 g
3 x 74.1 g Ca(OH) (^2) 6 x 18.0 g H2O
(1) the theoretical yield of Ca(OH) 2 is 30.0 g. ( 2) the limiting reactant is Ca 3 N 2.
(3) the mass of H 2 O reacted = 20.0 g Ca3N2 x (^) 148.3 g Ca3N2 = 14.6 g
6 x 18.0 g H2O
mass of H 2 O left over = (16.0 - 14.6) g = 1.4 g
(b) For the reaction 4PH 3 + 8O 2 H P 4 O10 + 6H 2 O calculate ( 1) the theoretical yield
of P 4 O 10 , and (2) the mass of the excess reactant left over when 35.0 g of PH 3 were reacted with 60.0 g of O 2. [ Ans. (1) 66.6 g P 4 O10, (2) 3.1 g PH 3 ]
(c) Consider the reaction 3NaBH 4 + 4BF 3 H 3NaBF 4 + 2B 2 H
The reaction was carried out using 78.0 g of NaBH4 and 172.0 g of BF 3. Calculate ( 1) the theoretical yield of B 2 H 6 , and ( 2) the mass of the excess reactant left over. [ Ans. (1) 35.0 g B2H6, (2) 6.1 g NaBH 4 ]
5. STOICHIOMETRY AND THERMOCHEMICAL EQUATIONS
In all chemical reactions, heat is either released (or evolved, or liberated) ( exothermic reactions ) or absorbed (or used) ( endothermic reactions ). When heat is released, heat can be treated (at the CHEM 094 level; a more rigourous treatment is used at the CHEM 105 and higher levels) as a product of the reaction. When heat is absorbed, it must be supplied and can be treated as a reactant in the reaction. An example of an exothermic reaction is the combustion of methane (CH 4 ; the main constituent of natural gas). The equation for the reaction is given below.
CH 4 ( g ) + 2O2( g ) H CO 2 ( g ) + 2H 2 O( l ) + 890 kJ
An example of an endothermic reaction is the reaction of nitrogen with oxygen to produce N 2 O. This reaction can be written
2N 2 ( g ) + O2( g ) + 163 kJ H 2N 2 O( g )
Chemical equations which include the amount of heat released or absorbed are called thermochemical equations and must have the physical states of all reactants and products specified. In these equations, the amount of heat released or absorbed are for the number of moles of reactants and products given in the equation. For example, when 1 mole of CH 4 ( g ) reacts with O 2 ( g ) to produce 1 mole of CO 2 ( g ) and 2 moles of H2O( l ), 890 kJ of heat are released. [Note that the reaction of 1 mole of CH 4 ( g ) with O 2 ( g ) to produce 1 mole of CO 2 ( g ) and 2 moles of H 2 O( g ) would produce less heat because heat is needed to convert H 2 O( l ) (liquid water) to H 2 O( g ) (water vapour)]. Hence, thermochemical equations can be treated as other equations for stoichiometric calculations.
Example The reaction for the combustion of ethane gas (C 2 H 6 ) is given below.
2C 2 H 6 ( g ) + 7O2( g ) H 4 CO 2 ( g ) + 6H 2 O( l ) + 3119 kJ
(a) Calculate how much heat is released (given off) when 555 g of ethane are reacted.
555 g C 2 H 6 x = 2.89 x 10^4 kJ 30.0 g C 2 H
1 mole C 2 H 2 moles C 2 H
x 3119 kJ
(b) Calculate how much water is produced when 955 kJ are released.
955 kJ x = 33.1 g H (^2) 3119 kJ
x
6 moles H 2 O 1 mole H 2 O
18.0 g H 2 O
