Asked by Jefferson
#5 Limiting Reactants
How many grams of water will be produced in the complete combustion of 20.0 grams of Octane, C8H18(g) and 50.0 grams Oxygen gas?
How many grams of water will be produced in the complete combustion of 20.0 grams of Octane, C8H18(g) and 50.0 grams Oxygen gas?
Answers
Answered by
DrBob222
This is a limiting reagent problem just like the ones that Devron worked for you this morning.
2C8H18 + 25O2 ==> 16CO2 + 18H2O
That should get you started.
Step 1. convert grams to mols. mols = grams/molar mass
Step 2. Using the coefficients in the balanced equation, convert mols of each to mols of the H2O.
Step 3. The smaller number of mols will be the correct value to use for mols H2O produced.
Step 4. Convert to g. g = mols H2O x molar mass H2O.
Post your work if you get stuck.
2C8H18 + 25O2 ==> 16CO2 + 18H2O
That should get you started.
Step 1. convert grams to mols. mols = grams/molar mass
Step 2. Using the coefficients in the balanced equation, convert mols of each to mols of the H2O.
Step 3. The smaller number of mols will be the correct value to use for mols H2O produced.
Step 4. Convert to g. g = mols H2O x molar mass H2O.
Post your work if you get stuck.
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