Sodium and chlorine react to form sodium chloride: 2na(s)+cl2(g)→2nacl(s)part awhat is the theoretical yield of sodium chloride for the reaction of 55.0 g na with 67.2 g cl2?

111 grams First, look up the atomic weights of sodium and chlorine. Atomic weight sodium = 22.989769 Atomic weight chlorine = 35.453 Molar mass Cl2 = 2 * 35.453 = 70.906 g/mol Molar mass NaCl = 22.989769 + 35.453 = 58.442769 g/mol Now determine how many moles of each reactant you have. Moles Na = 55.0 g / 22.989769 g/mol = 2.392368536 mol Moles Cl2 = 67.2 g / 70.906 g/mol = 0.947733619 mol Looking at the balanced equation, it takes 2 moles of Na per mole of Cl2. Given the number of respective moles, it's obvious that Cl2 is the limiting reactant in this problem. And looking at the balanced equation, for every mole of Cl2 consumed, 2 moles of NaCl is produced. So do the multiplication. 2 * 0.947733619 mol = 1.895467238 mol And multiply by the molar mass, so 1.895467238 * 58.442769 = 110.776354 g Round to 3 significant figures, giving 111 g

Hbr dissociates to form h+ and br-ca(oh)2 dissociates to form ca2+ and 2oh-the h+ and oh- react to form waterassuming the reactants are dissolved in water, the net reaction is h+ + oh- —> h2oif you’re asked to show all components of the reaction, the complete reaction is 2hbr + ca(oh)2 —> cabr2 + 2h2o