Feb 01, 2012 · HELP PLEASE!!! 1. What is the mole ratio of O2 to H2O in the reaction C6H12O6 + 6O2 --> 6CO2 + 6H2O? 2. What kind of reaction is shown in question 1 above? 3. How many grams of O2 will be produced if 46 moles of O3 are in the reaction O2--> O3? 4. What is the theoretical yield of nitrogen if 3.8 mol of nitroglycerin (C3H5N3O9) is detonated according to the following reaction? C3H5N2O9 --> O2 ...
• Converting the masses to moles of elements leads to the empirical formula, the simplest whole-number ratio of moles of each element in the compound. Plan: -Use changes in mass of CO 2 and H 2O absorbers to calculate the moles of C and H present in the sample.
molecules or moles of participants (reactants and products) in a chemical reaction. To be valid, the equation must be balanced. For example, the following stoichiometric equation is not balanced: C2H5OH+O2 →CO2 +H2O • The following equation is balanced because the number of atoms is the same on both sides of the equation (C, H, and O):
Mole Ratio Worksheet Mole Ratio Worksheet 1) ... ratios: a) H2 / H2S b) H2 / S8 c) H2S / S8 3) Answer the following questions for this equation: 2 H2 + O2---u0026gt; 2 H2O a) ... [Filename: WS1-MoleRatio.pdf] - Read File Online - Report Abuse
2c4h10 13o2=8co2 10h2o write the mole ratio of c4h10 to o2. Unknown008 Posts: 8,076, Reputation: 723. Uber Member : Jan 12, 2011, 11:00 PM ...
means moles of H2O:CO of 1:1, or 2:2, etc. Compute the mole fractions of all components assuming the extent of reaction is, ξ, 0.25. Also compute the ratio of product mole fractions to reactant mole fractions. (e) Reconsider the reaction CO + H2O = CO2 + H2, assuming a feed in which the moles of H2O and CO are 2 and 1.
Therefore, the equation needs a total of 12 O2s (2 O2 for the hydrogen + 10 O2 for the carbon). Thus, the equation becomes so: 1 C10H8 + 12 O2 -----|> 10 CO2 + 4 H2O. This equation reads: 1 mole of naphthalene combine with 12 moles of diatomic oxygen to produce 10 moles of carbon dioxide and 4 moles of hydrogen monoxide (water).