Chemistry Stoichiometry Problem: Sodium and Chlorine Reaction
How many grams of sodium chloride (salt) are formed if 23 g of sodium reacts with 71 g of chlorine?
Final answer:
23 grams of sodium reacting with 71 grams of chlorine will produce 58.44 grams of sodium chloride, using the stoichiometry of the balanced chemical equation and the molar masses of the reactants.
Answer:
The problem involves a chemical reaction between sodium and chlorine to produce sodium chloride, which is a classic stoichiometry problem in chemistry. We can solve this by using the law of conservation of mass and the stoichiometric coefficients from the balanced chemical equation 2 Na(s) + Cl2(g) → 2 NaCl(s). This equation tells us that for every mole of chlorine gas that reacts, two moles of sodium are required to form two moles of sodium chloride.
First, we must calculate the moles of sodium and chlorine. One mole of sodium (Na) has a mass of 22.99 grams, so 23 grams is essentially 1 mole of Na. One mole of chlorine (Cl2) has a mass of 70.90 grams, so 71 grams is slightly more than 1 mole of Cl2.
According to the balanced equation, 2 moles of Na react with 1 mole of Cl2 to produce 2 moles of NaCl. Since sodium is the limiting reactant (we only have 1 mole of Na), the reaction will produce 1 mole of NaCl (because we need 2 moles of Na for the reaction to use up all the chlorine). The molar mass of NaCl is 58.44 grams/mole, so 1 mole of NaCl weighs 58.44 grams.
Therefore, 23 grams of sodium reacting with 71 grams of chlorine will produce 58.44 grams of sodium chloride.