1.16g of methane burns completely with 4.16g of oxygen to form 3.52g of carbon dioxide and water If the reaction obeys the law of cons

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1. Answer:

   weight of H₂O formed = 2.3 grams H₂O (2 sig. figs.)

   Explanation:

   Rxn:                  CH₄(g)         +      2O₂(g)          =>     CO₂(g)           +  2H₂O(g)

   Given:                 1.16g                   4.16g                     3.52g                  ? (g)

   Moles:         1.16g/16g·mol⁻¹    4.16g/32g·mol⁻¹    3.52g/44g·mol⁻¹

                        = 0.0725mol.        = 0.13mol.             = 0.08mol.    => ? (moles)

   Limiting Reactant: Divide each mole value by related coefficient of balanced standard equation (that is, balanced with coefficients in lowest whole number ratios). The smaller value is the limiting reactant.

                       0.0725/1                   0.13/2                0.080/1

                       = 0.0725                 = 0.065              = 0.080

   Limiting Reactant is O₂(g) => 0.065 is smaller value after dividing each mole value by related coefficient of balanced equation.

   NOTE: When working problem, however, one must use the mole value calculated from given amount in grams. That is, in this case 0.13 mole O₂. The ‘divide by related coefficient and check smaller value’ is ONLY for identifying the limiting reactant. This trick works for ALL general chemistry problems.

   Moles H₂O formed: Since the coefficient of the limiting reactant (O₂) equals the coefficient of water (H₂O), then the moles of water formed is 0.065 mole H₂O.

   Weight (in grams) of H₂O formed:

   Grams H₂O = moles H₂O  x  formula weight H₂O

                      = 0.13 mole H₂O  x  18 g H₂O/mole H₂O

                      = 2.34 g H₂O  (calculator answer)

                      = 2.3 g H₂O  (final answer should be rounded to  2 sig. figs.)                                                                                                                                        => form of final answer should be based on data in final computation having the least number of sig. figs.                                      

   Review: Sequence of calculations

   • Write and balance equation to smallest whole no. ratio of coefficients.
   • If not in moles, convert given ‘measured’ data to dimension of moles.    => moles = mass (g)/formula wt(g·mol⁻¹)                                                   => moles = volume of gas in Liters/Std Molar Volume (= 22.4L·mole⁻¹ at STP)                                                                                                               => moles = no. of particles / Avogadro’s No. (= 6.02 x 10²³ part’s/mole)
   • Determine Limiting Reactant => mole values of each compound given / related coefficient in standard equation => smallest value is L.R.
   • Determine moles of unknown needed/used/formed from limiting reactant in moles and coefficient of unknown compound in standard equation given data values.

          =>  moles of limiting reactant / coefficient of same cpd. in std. equation               = unknown (X) / coefficient of same (unknown) cpd. in std. equation

          => cross multiply and solve for unknown (X)

          => L.R.(moles) / eqn. coef. of L.R. = X / eqn. coef. of X  

          => (L.R.(calc’d moles)(eqn. coef. of X) = (X)(eqn. coef. of L.R.)

          => X (in moles) = (L.R.(calc’d. moles)(eqn. coef. of X) / (eqn. coef. of L.R.)

   • Convert X-answer in moles to desired dimension specified in problem.    

           => grams = moles x formula wt.                                                                

           => volume (L) = moles x std. volume (= 22.4L/mole)                                  

           => #particles = moles x Avogadro’s Number (= 6.02 x 10²³ parts/mole)

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