mole in stoichometry.

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Explanation: 

(a)        How many moles of oxygen would be needed to react 58.5 grams of H2 to form       water?

      Solution;

      Mass of hydrogen = 58.5 grams

      Moles of hydrogen 58,5/ 2.016 = 29.02 moles

      According to balance equation,

      2 moles hydrogen need oxygen         = 1 mole

      29.02 moles hydrogen need oxygen  = 29.02/ 2 = 14.51 moles of oxygen is         required for 58.5 grams of hydrogen

(b)        How many grams of H2 would be needed to form 120 grams of water?

                        Solution;

                        Mass of water obtained = 120 grams

                        Moles of water obtained  120/ 18 = 6.67 moles

                        Now according to balance equation,

                        2 moles H2O need H2          = 2 moles

                        6.67 moles H2O need H2  = 2x 6.67/ 2 = 6.67 moles of H2 is required

                        Grams of hydrogen required = mole x M. weight

                                                                          = 6.67 X 2.016 = 13 .44 grams of hydrogen is required

 

 

Q                Given the equation                        N2 + 3H2 → 2NH3 at STP. How many moles of NH3                           would be formed, If 6.3 dm3 of N2 gas react with an excess of hydrogen.?

                        Solution;

                        Volume of I mole of a gas at stp = 22.414 dm3

                        Volume of Nitrogen given is= 6.3 dm3

                        Moles of nitrogen                 = Volume in dm3      = 6.3 dm3       = 0.28125  moles

                                                                           Molar volume            22.414 dm3

                        According to balance equation ,

                        1 mole of N2 give      NH3                 =  2 moles

                        0.28125 moles of N2 give NH3  = 2x 0.28125 = 0.5625 moles of NH3 will obtain

                        1 mole nitrogen give NH3 = 2 mole

 

Q               Calculate the mass of Mg metal required to consume 2560 grams of CO2 in                                        reaction.        2Mg(g) + CO2(g) → 2MgO(s) + C(s)

                        Solution;

                        Mass of CO2 = 2560 grams

                        Moles of CO2 = mass of CO2= 2560   = 58.18 moles

                                                   Molar mass         44

                        According to the balance equation ,

                        1 mole CO2  need Mg         = 2 moles

                        58.18 moles CO2 need Mg =   x =   2x 58.18 = 116.36 moles of Mg is required

                        Mass of Mg required is = Mole x molar mass = 116.36 x 24 =  2792.73 grams

 

Q                when steam is passed through red hot carbon., a mxture of H2 and CO gases                                   called water gas is formed.

                        C(s) + H2O(g) → CO(g) + H2(g)

(a)        Which is the limiting reagent if 24.5 grams of carbon is mixed with 1.89 moles of water       vapors?

      Mass of carbon = 24.5 g

      Mole of carbon 24.5/12 = 2.04 moles

      Mole of water = 1.89 moles

      As one mole of water need one mole of carbon, therefore 1.89 moles of water      need 1.89 moles of C and we have 2.04 moles of C. therefore water is limiting       reactants. While carbon is reactants in excess.

 

(b)        Calculate the amount ( in grams) of the excess reagent left unreacted.

                        Total amount of carbon = 2.04 mole

                        Carbon consumed        = 1.89 moles

                        Carbon remained unreacted; 2.04 – 1.89 =  0.15 moles

                        Mass in grams of carbon = mole x molar mass= 0.15 x 12=  1.8 grams