1) Consider the unbalanced equation for the combustion of hexane: 2C6H14(g)+19O2

Question

1) Consider the unbalanced equation for the combustion of hexane:
2C6H14(g)+19O2(g)12CO2(g)+14H2O(g)
Determine how many moles of O2 are required to react completely with 6.8 moles C6H14.

2) CH4(g)+2O2(g)CO2(g)+2H2O(g)

A)What mass of carbon dioxide is produced from the complete combustion of 1.20×103 g of methane?
Express your answer with the appropriate units.

B)What mass of water is produced from the complete combustion of 1.20×103 g of methane?
Express your answer with the appropriate units.

C) What mass of oxygen is needed for the complete combustion of 1.20×103 g of methane?
Express your answer with the appropriate units.

3) Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange and pineapple. Its fragrance and taste are often associated with fresh orange juice, and thus it is most commonly used as orange flavoring. It can be produced by the reaction of butanoic acid with ethanol in the presence of an acid catalyst (H+):CH3CH2CH2CO2H(l)+CH2CH3OH(l)H+CH3CH2CH2CO2CH2CH3(l)+H2O(l)

A) Given 7.20 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield?
B) A chemist ran the reaction and obtained 5.15 g  of ethyl butyrate. What was the percent yield?
C) The chemist discovers a more efficient catalyst that can produce ethyl butyrate with a 78.0% yield. How many grams would be produced from 7.20 g of butanoic acid and excess ethanol?

4) Melamine, C3N3(NH2)3, is used in adhesives and resins. It is manufactured in a two-step process in which urea, CO(NH2)2, is the sole starting material, isocyanic acid, HNCO, is an intermediate, and ammonia, NH3, and carbon dioxide, CO2, gases are byproducts.

In the first step of this reaction, urea is broken down into isocyanic acid and ammonia:
CO(NH2)2(l)HNCO(l)+NH3(g)

In the second step of this reaction, isocyanic acid reacts to form melamine and carbon dioxide:
6HNCO(l)C3N3(NH2)3(l)+3CO2(g)

A)What mass of melamine, C3N3(NH2)3, will be obtained from 157.5 kg of urea, CO(NH2)2, if the yield of the overall reaction is 72.0 % ?
Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

1. 2C6H14 + 19O2 ---> 14H2O + 12CO2
You need 19 moles oxygen for every 2 moles of hexane you burn:
6.8 x 19/2 = 64.6 moles

2. Equation: CH4 + 2O2 CO2 +2 H2O
1mol CH4 will produce 1 mol CO2
Molar mass CH4 = 16g/mol
1.20*10^-3 g CH4 = (1.20*10^-3) / 16 = 7.5*10^-5 mol CH4
You will produce 7.5*10^-5 mol CO2
Molar mass CO2 = 44g/mol
7.5*10^-5 mol = 44 * (7.5*10^-5) = 0.0033g CO2 produced OR 3.3*10^-3g. CO2 produced.

Second problem : 1mol CH4 will produce 2 mol H2O
1.20*10^-3 mol CH4 will produce 2.4*10^-3 mol H2O
Molar mass H2O = 18g/mol
2.4*10^-3 mol H2O = 18*2.4*10^-3 = 0.0432 g H2O produced

Third problem:
1 mol CH4 will react with 2 mol O2
1.20*10^-3 mol CH4 will react with 2.4*10^-3 mol O2
Molar mass O2 = 32g/mol
2.4*10^-3 mol O2 = 32*2.4*10^-3 = 0.0768g O2 reacted

3. Chemical Rxn: H7C3-COOH + OH-C2H5 -> H7C3-COO-C2H5 + H2O

1) [butanoic acid, C4H8O2] = 48 + 8 + 32 = 88
No. of moles of butanoic acid used = 7.20 / 88 = 8.18 x 10^-2 mol

Since stoichiometric ratio of butanoic acid and it's esterification product is 1:1, AND assuming 100% yield,
No. of moles of ethyl butyrate = No. moles of butanoic acid = 8.18 x 10^-2 mol * 88 = 7.19 g

2) 5.15/7.19 x 100% = 71.62%

3) 78% = 78/100
78/100 x 7.19 = 5.60g

4. CO(NH2)2(l) HNCO(l) + NH3(g)
6 HNCO(l) = C3N3(NH2)3(l) + 3 CO2(g)

(157.5 kg CO(NH2)2) / (60.05 g CO(NH2)2/mol) x (1 mol HNCO / 1 mol CO(NH2)2) x
(1 mol C3N3(NH2)3 / 6 mol HNCO) x (126.12 g C3N3(NH2)3/mol) x (0.750) = 39.69 kg C3N3(NH2)3

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