Law of Conservation of Mass Chemistry Lab

Balancing chemical equation is the most important part of the Law of conservation of mass. It’s used to make sure the mass of initial reactants matches the mass of the products: in other words, the mass of any one element at the beginning of a reaction will equal the mass of that element at the end of the reaction. To balance a chemical equation, we need to follow these steps. First, write down the number of atoms per element before and after the reaction. Second, save the H and O for the end of the procedure. Next use stochiometric coefficient to balance the singled-out element (the element that appears in only a single molecule of reactants and in only a single molecule of product). Then use stochiometric coefficient to balance H and finally balance the O’s. In this lab we had to use a simulation call PHET to balance out chemical equations. We’ll discuss about prior chemical equations that we earlier balanced (one reaction from the Easy category, one from the Medium, and one from the hard). Our work will then consist into explaining the procedure different steps and relate our findings to the law of conservation of mass through the molar mass. The introduction simulation shows how ammonia is made. At the beginning of this simulation the initial reactants N2 and H2 and their product NH3 have 0 for stochiometric coefficient. First, we identified the single-out element N. Next, we clicked upward 2 times to balance out N2 and NH3. Finally, we balance out the H’s with 3clicks upward giving us a fully balanced equation: Before: N2 After: + H2  NH3 2 N2 + 3 H2  2 NH3 After balancing out the equation we need to check that the sum of the reactants molar mass is equal to the molar mass of the product 2 N2 + 3 H2 = (2 * 14.00) + (3 x 2 x 1.001) = 36.006 g/mol 2NH3 = 36.006 g/mol When we opened game on the simulation, we had to work on three different levels of equations: easy, medium and, hard. We used stoichiometry to balance each chemical equation. Here are some the equations the simulation offered: Easy: Before: 0N2 + 0O2  0N2O After: 2N2 + 1O2  2N2O Molar mass: (2 x 14.00) + (1 x 2x 16.00) = (2 x 14.00) + (1 x 16.00) 42.00 g/mol = 42.00 g/mol At the beginning of the simulation the equation alle the elements and compound have 0 for stochiometric number. Click twice on the upward arrow to balance N 2, once for O 2 and, twice for N2O. Next, we summed the molar masses of the reactants and it’s equal to the product’s molar mass. The equation is then balanced. Medium: Before: 0SO2 + 0H2  0 H2S + 0H2O After: 1SO2 + 3H2  1H2S + 2H2O Molar mass: [(1 x 36.07) +(2 x 16.00)] + [3 x 2 x 1.01]  (1 x 2 x 1.01) +(1 x 36.07) + (2 x 2 x 1.01) + (1 x 16.00) 74.13 g/mol = 74.13 g/mol In the second part of the simulation, the equation all the elements and compound have 0 for stochiometric number. First, we balance reactants; we click once upward to balance SO 2 and 3 times upward to balance H2. Next, we balance out the reactants we clicked upward 1 time to balance H2S and 2 times for H 2O. Next, we summed the molar masses of the reactants and it’s equal to the product’s molar mass. The equation is then balanced. Hard: Before: 0 NH3 + 0 O2  0 NO2 + 0 H2O After: 4 NH3 + 7 O2  4 NO2 + 6 H2O Molar mass: [(4 x14.00) +( 4 x 3 x 1.008)] + (7 x 2 x 16.00) = [(4 x 14.00) + (4 x 2 x 16.00)] + [(6 x 2 x 1.008) + (1 x 16.00)] 292.12 g/mol = 292.12 g/mol In the last part of the simulation, the equation all the elements and compound have 0 for stochiometric number. First, we balance reactants; we click 4 times upward to balance NH 3 and 7 times upward to balance O2. Next, we balance out the reactants we clicked upward 4 times to balance NO2 and 6 times for H2O. Next, we summed the molar masses of the reactants and it’s equal to the product’s molar mass. The equation is then balanced. The more we look at these equations we realize other using stoichiometry to balance the equation we can also use their molar. It corelates with Lavoisier’s thoughts on the law of conservation of mass that mass is neither created nor destroyed in chemical reactions. In other words, the mass of any one element at the beginning of a reaction will equal the mass of that element at the end of the reaction.