Question

Assume a reaction where magnesium (Mg) reacts with chlorine (Cl2) to produce magnesium chloride (MgCl2). If we start with 10 grams of magnesium and 20 grams of chlorine, what mass of magnesium chloride can we produce?

Answer 1

To solve this problem, we will use the concept of limiting reactants and stoichiometry. First, we need to determine the balanced chemical equation for the reaction:Mg + Cl2  MgCl2Next, we need to find the molar masses of the reactants and product:Mg: 24.31 g/molCl2: 2 * 35.45 g/mol = 70.90 g/molMgCl2: 24.31 g/mol + 2 * 35.45 g/mol = 95.21 g/molNow, we will convert the given masses of reactants to moles:Moles of Mg: 10 g / 24.31 g/mol = 0.411 molesMoles of Cl2: 20 g / 70.90 g/mol = 0.282 molesTo determine the limiting reactant, we will compare the mole ratios of the reactants:Mole ratio of Mg to Cl2: 0.411 moles / 0.282 moles = 1.46Since the balanced chemical equation has a 1:1 ratio for Mg and Cl2, and the mole ratio of Mg to Cl2 is greater than 1, Cl2 is the limiting reactant.Now, we will use the stoichiometry of the balanced chemical equation to determine the moles of MgCl2 produced:Moles of MgCl2 = moles of limiting reactant  Cl2  = 0.282 molesFinally, we will convert the moles of MgCl2 to grams:Mass of MgCl2 = moles of MgCl2 * molar mass of MgCl2 = 0.282 moles * 95.21 g/mol = 26.84 gTherefore, we can produce 26.84 grams of magnesium chloride.