InterviewSolution
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InterviewSolution
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What mass of oxygen gas is required to burn completely `2.5 mol` of methane? `underset(underset("reactant")("one"))("Moles of ")rarr underset(underset("reactant")("another"))("Mass of")` Stragetgy : Use the balanced equation `{:(CH_(4), + 2O_(2), rarr CO_(2), + 2H_(2) O),(1mol, 2mol,1 mol,2mol),(16.0 g,2(32.0)g,44.0g,2(18.0g)):}` to find the relationship among moles and grams of reactants : `underset(CH_(4))("Mole of ")rarr underset(O_(2))("Moles of")rarr underset(O_(2))("grams of")` |
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Answer» We can solve the problem in three steps. Step1: Mass `CH_(4) rarr mol CH_(4)` `? "mol" CH_(4) = 40.0g CH_(4) xx (1 mol CH_(4))/(16.0g CH_(4)) (n = ("Mass")/("Molar mass"))` `= 2.50 "mol" CH_(4)` Step 2 : `"mol" CH_(4) rarr mol O_(2)` `? "mol" O_(2) = 2.50 "mol" CH_(4) xx (2 "mol" O_(2))/(1 "mol" CH_(4))` (Balanced equation used) `= 5.00 "mol" O_(2)` Step 3 : mol `O_(2) rarr` mass `O_(2)` `? g O_(2) = 5.00 "mol" O_(2) xx (32.0 g O_(2))/(1 mol O_(2)) ("Mass = n xx MM")` `= 160g O_(2)` Alternaty all these steps can be combined into one setup as follows : `underset(CH_(4))("g of")rarrunderset(CH_(4))("mol of")rarr underset(O_(2))("mol of")rarr underset(O_(2))("g of")` `? g O_(2) = 40.0g CH_(4) xx (1 "mol" CH_(4))/(16.0g CH_(4)) xx (2 "mol" O_(2))/(1 "mol" CH_(4))` `xx (32.0 g O_(2))/("1 mol" O_(2))` `= 160g O_(2)` |
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