When working stoichiometric problems in an all gas system, one may use a shortcut in which volume may be used as mols directly. Therefore, all you need to do is to convert L of one material to L of the other; it is not necessary to go through mols.
For example:
7.70L BNH x (15 L O2/4 L BNH) = ? L O2 needed.
and
7.70 L BNH x (8 mols CO2/4 mols BNH) = ? L CO2 formed.
Note: It was convenient that the CONDITIONS (0.840 atm and 35 C) did NOT change. Had they changed we would have been forced to use PV = nRT to obtain mols.
Dimethylamine burns in oxygen according to the following equation:
4 C2H7N + 15 O2 → 8 CO2 + 14 H2O + 2 N2
(a) How many liters of O2 at 35 °C and 0.840 atm will be needed to burn 7.70 L of C2H7N at 35 °C
and 0.840 atm?
(b) How many liters of CO2 at 35 °C and 0.840 atm will be produced?
Report your answers to parts (a) and (b) to 3 significant figures
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