ENPE_381_Set-1-2010-Solution

Question 13 of 30 Submit A 7.591-g gaseous mixture contains methane (CH4) and butane (C,H10). Complete combustion to form carbon dioxide and water requires 0.9110 mole of oxygen gas. Calculate the mass percent of methane in the original mixture. % 1 4 6. C 7 +/- x 10 0 Tap here or pull up for additional resources 3. LO 00

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observe the following balanced chemical equation: 6 NH3(g) + 8 O3(g) --> 3 N2O5(g) + 9 H2O(g) if the reaction of 2.0 mol NH3(g) and 2.5 mol of O3(g) is run with 100% efficiency which of the followng is correct?   all NH3(g) will be consumed 60.0 g of water will be produced all of the O3(g) will be consumed 1.0 mole of N2O5(g) will be produced

In one sentence explain what's wrong with this equation, and rewrite it so that the coefficients are correct.  C3H6(g)+4.5 O2(g)->3CO2(g)+3H2O(g)

12. What is the mole ratio of oxygen to carbon dioxide in the combustion reaction for propane? C3Hg+502 3CO2 + 4H20 5:3 5:1 3:5 O1:5 E 00:54:42 PREVIOUS 12 of 12 SUBMIT ALL A. %23 W Y u 5

Balance the following chemical equation (ifnecessary):C6H6O(I) + O2(g) -> H20 (g) + CO2 (g)

A compound contains, by mass, 81.7% carbon and 18.3% hydrogen. A 0.15 mole sample of this compound weighs 6.6 g. What is the molecular formula of this compound? C2H6 CH3 C3H12 C4H10 C3H8

how do you solve  ______C10H14N2(s)+ ______O2(g)---->____CO2(g)+____H2O(l)+_____NO2(g)?

How many grams of N2 are needed to make 444 g of NBr3 ?  N2   +   3 Br2   →   2 NBr3   +   44.0 kJ              The molar mass of NBr3 is 253.719 g and the molar mass of N2 is 28.013 g                        _______________ mol NBr3                                                                                   (444 g NBr3)(–––––––––––––––––––––––––)  = ______________  mol NBr3                        ____________________ g NBr3                                                                                                                                          _______________ mol N2    _______________ mol NBr3 (–––––––––––––––––––––––––) =  ___________---- mol N2                                                        _________________ mol NBr3                                                            _______________  g N2 _____________________ mol N2 (–––––––––––––––––––––––––) = ___________________  g N2                                                       __________________ mol…

A local water retention pond was found to have elevated levels of mercury concentrated as 14 ng per liter. You would like to collect some mercury for your homemade perpetuum mobile machine. How many moles of mercury can you get you collect and process the entire volume of 1250 m3 of the water in the pond? 1.8 x 10-5 moles 4.4 x 10:5 moles 1.8 x 10-2 moles 8.7 x 108 moles 7.6 x 10-5 moles O 4.4 x 108 moles 8.7 x 105 moles

Write a balanced chemical equation based on the following description: solid CHO is burned with oxygen gas to 10 produce gaseous carbon dioxide and water vapor 2 C₂²H₂O² H(aq) + 9 O₂(g) →6 C4O₂(g) + 8 H₂O²- (1)

9. A mixture of 50 g H₂0 and 30 g of C6H5OH separates into two layers at 62°C. One layer, X₁, consists of 46.2% H₂O by mass; the other X2, consists of 82.7% H₂O by mass. b. What are the masses of X₁ and X₂?

Sodium hydroxide reacts with carbon dioxide as follows: 2NaOH(s)+CO2(g)→Na2CO3(s)+H2O(l)   How many moles of Na2CO3 can be produced?

9. A mixture of 50 g H₂0 and 30 g of CeH₂OH separates into two layers at 62°C. One layer, X₁, consists of 46.2% H₂O by mass; the other X₂, consists of 82.7% H₂O by mass. c. What are the total number of moles in X₁ and X₂?

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T Ma M PA USN Be * Ma E Ad G ph ph Do E Un O Ac. 6 Pec O Na O Pe Со Bas 3 Pe Stu "app.101edu.co Question 11 of 11 What is the mass in grams of H2 that can be formed from 55.8 grams of NH3 in the following reaction? 2 NH:(g) → 3 H:(g) + N:(g)

complete combustion CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) + heat incomplete combustion 6 CH4(g) + 9 O2(g) → 2 C(s) + 2 CO2(g) + 12 H2O(l) + 2 CO(g) + heat

1.    Water is formed by burning H2 (g) in the presence of O2 (g):    2H2 (g) + O2 (g) à 2H2O (g) a)    If you have 32 g of H2­ gas and 32 g of O2­ gas, which do you have the most of in terms of moles? Explain your reasoning.   b)    Starting from the 32 g of H2, how many moles of water can form from the H2­ gas?     c)    Starting from the 32 g of O2, how many moles of water can form from the O2­ gas?   d)    Based on the amounts of water formed in parts b) and c), which is the limiting reactant for this reaction? H2­ gas or O2­ gas? Explain your reasoning.   e)    According to the limiting reactant you chose in part d), what is theoretical yield in grams of water that could be produced when 32 g of H2­ gas reacts with 32 g of O2­ gas?

3. Using this data, calculate a value for Kc for the reaction, A. 2.06 x 104 B. 4.84 x 10-3 C. 0.223 D. 4.49 E. 20.2 2 NO(g) + Cl₂(g) 2. NOCI(g) Kc = 3.20 x 10-³ 2 NO2(g) 2 NO(g) + O₂(g) Kc = 15.5 4. Using the data, NOCI(g) + 1/2O2(g) NO2(g) + ½ Cl₂(g) 3 A (g) + 4 B (g) A3B4 (g) calculate a value for Kc for the reaction, 3 A (g) + 4 B -A3B4 (g) 2 AB (g) + AB2 (g) (g) Kc = 25 Kc = 3.75 2 AB (g) + AB2 (g)

Consider the following reaction for formation of nitric acid from the pollutant nitrogen dioxide reacting with oxygen and water in the atmosphere.   4 NO2 (g) + O2 (g) + 2 H2O(l) ® 4 HNO3 (aq) The generation of electricity used by a medium-sized home produces about 16 kg of NO2 per year. What mass of nitric acid, in kg, can form from this amount of NO2 pollution? Assume oxygen and water are excess reactants.

To help farmers and gardeners , commercial fertilizers have a big three -number "NPK" label on the bag that gives the amounts of three key plant nutrients: nitrogen , phosphorus and potassium. For example , a bag of "-10-10-10" fertilizer has 10% by mass nitrogen , 10% by mass potassium as potash (K2O), and 10% by mass phosphorus as phosphate (P2O5). Suppose a certain fertilizer has the following composition. What should the first (nitrogen) number on the bag label be? That is, calculate the the percent by mass of nitrogen in this fertilizer. Round your answer to the nearest percent.

: ny F1 ! Q What does a complete balanced chemical equation include? O F2 molar coefficients valence electrons exothermic coefficients 2 endothermic coefficients -0- @ W F3 -Ő+ #m 3 E F4 $ 4 F5 R % F6 T Q F7 6 Y F8 & 7 F9 *00 U 8 H

The ORSAT analysis of a flue gas is: Note: 100 mole of flue gas..                                         CO2 = 12.0%;                O2 = 8.0%;             N2 = 80.0% If the nitrogen present in the flue gas is contributed by only air, then what is the percent of excess air used in the combustion?

The molecular formula and the empirical formula for this hydrocarbon are a. C5 H10 and CH2 b. C5 H10  and CH3 c. C4H8 and CH2 d. C4H8 and CH3