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 conservation of mass, the weight of water formed is

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

       =>  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.)

        => 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)

personal, she chose to do it.

the answer is 0.8m.

by: nicholas