Exam-2-Solution

# Exam-2-Solution - 1 The following balanced equation for the...

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Unformatted text preview: 1. The following balanced equation for the combustion of octane tells us which one of the following: (2 pts) 2 C 8 H 18 ( l ) + 25 O 2 ( g ) → 16 CO 2 ( g ) + 18 H 2 O ( g ) Answer : That 25 mols of O 2 ( g ) are consumed for every 2 mols of C 8 H 18 ( l ) consumed. 2. Using the chemical equation from problem 1, if 10.0 mols of C 8 H 18 ( l ) are allowed to react in an excess of oxygen, how many mols of H 2 O ( g ) will be produced? (2 pts) Answer : 10 mols C 8 H 18 × 18 mols H 2 O 2 mols C 8 H 18 = 90.0 mols H 2 O 3. Using the chemical equation from problem 1, if 10.0 g of C 8 H 18 ( l ) are allowed to re- act in an excess of oxygen, what is the mass ( in grams ) of CO 2 ( g ) will be produced? The molar mass of octane is 114.23 g/mol. (4 pts) Answer : 10.0 g C 8 H 18 × 1 mol 114 . 23 g × 16 mols CO 2 2 mols C 8 H 18 × 44 . 01 g 1 mol = 30.8 g CO 2 4. In the following equation describing the combustion of methanol, a, b, c and d represent the coefficients in the balanced equation. Identify the proper values for these coefficients. (4 pts) a CH 3 OH ( g ) + b O 2 ( g ) → c CO 2 ( g ) + d H 2 O ( g ) Answer : a=2; b=3; c=2; d=4 5. Use the balanced equation from problem 4 to calculate the theoretical yield ( in grams ) of water if the initial amounts of reactants are 4.00 g of methanol gas and 10.0 g of oxygen gas. The molar mass of methanol is 32.04 g/mol. (5 pts) 4.00 g methanol × 1 mol CH 3 OH 32 . 04 g × 4 mols H 2 O 2 mols CH 3 OH × 18 . g water 1 mol water = 4.50 g H 2 O 10.00 g methanol × 1 mol O 2 32 . 00 g × 4 mols H 2 O 3 mols O 2 × 18 . g water 1 mol water = 7.5 g H 2 O The smaller mass of water is the amount coming from methanol, therefore that is the theo-...
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## This note was uploaded on 04/19/2011 for the course CHEM 1031 taught by Professor Thomas during the Fall '06 term at Temple.

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Exam-2-Solution - 1 The following balanced equation for the...

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