eatlanl drid hitrogen is in excess. gin guo, jou al 2mol of ämmonia. Here, you a

Question

eatlanl drid hitrogen is in excess. gin guo, jou al 2mol of ämmonia. Here, you are sh When you combust methane, you need 2 mol of oxygen gas for every mole of methane. If you have 5 mol of methane and 5 mol h methane is available, you do not have enough oxygen gas to produce more carbon dioxide. Thus, you can cor Part D Find the number of moles of water that can be formed if you have 190 mol of hydrogen gas and 90 mol of Express your answer with the appropriate units. oxygen gas Hints ValueUnits Submit My Answers Give Up O Type here to search

Explanation / Answer

Part-D.

2H2 + O2 -----2 H2O

number of molesof H2 =190 moles

number of moles of O2 =90 moles

A/c to equation ,

1mole of O2 = 2 mole of H2

90 moles of O2 = ?

                       = 2x90/1 = 180 moles of H2

we need 180 moles of H2. but we above 190 moles of H2. So it is excess reagent

Hence O2 is an limiting reagent

a/c to equation, 1mole of O2 = 2 moles of H2O

                          90 moles of O2 = ?

                                                        90x2= 180 moles of H2O is formed.

number of moles of H2O formed = 180 moles

Problem 7.2

CH4 + 2 O2 ------------------ CO2 +2 H2O

one mole of an contains Avagadro number of atoms ( 6.023x10^23)

number of atoms of Hydrogens atoms= 4x 6.023x10^23 atoms

number of atoms of carbon on reactatt = 1x6.023x10^23 atoms of oxyger

number of at0ms Ogygen on reactants side= 4x6.023x10^23 ATOMS OF Oxygen

number of atoms of Oxygen on products= 4 x 6.023 x10^ 23 atoms

number of atoms of carbon = 1x6.023x10^ 23 atomd of carnon on products side

number of hydrogen atoms i 4x6.023 X10623 atom of carbob atoms

                                  

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