Understand how to solve stoichiometry problems.

Example 1: Calculate the mass of sodium chloride produced when 5 grams of sodium reacts with excess chlorine gas.

Step 1: Write the balanced chemical equation for the reaction.

2Na + Cl2 → 2NaCl

Step 2: Calculate the moles of the limiting reactant.

n = m/M = 5 g / 23 g/mol (molar mass of sodium) = 0.217 mol

Step 3: Use the mole ratio from the balanced equation to calculate the moles of the product formed.

From the equation, 2 moles of NaCl are produced for every 1 mole of Na. Therefore, the moles of NaCl produced = 2 x 0.217 mol = 0.434 mol

Step 4: Convert the moles of product to mass.

mass = n x M = 0.434 mol x 58.44 g/mol (molar mass of NaCl) = 25.3 g

Therefore, 25.3 grams of sodium chloride are produced.

Example 2: How many grams of magnesium oxide are produced when 10 grams of magnesium reacts with 10 grams of oxygen gas?

Step 1: Write the balanced chemical equation for the reaction.

2Mg + O2 → 2MgO

Step 2: Calculate the moles of each reactant.

n(Mg) = 10 g / 24.31 g/mol (molar mass of Mg) = 0.411 mol

n(O2) = 10 g / 32 g/mol (molar mass of O2) = 0.3125 mol

Step 3: Determine the limiting reactant.

The reactant with the smaller number of moles is the limiting reactant. In this case, oxygen is the limiting reactant.

Step 4: Use the mole ratio from the balanced equation to calculate the moles of the product formed.

From the equation, 2 moles of MgO are produced for every 1 mole of O2. Therefore, the moles of MgO produced = 2 x 0.3125 mol = 0.625 mol

Step 5: Convert the moles of product to mass.

mass = n x M = 0.625 mol x 40.3 g/mol (molar mass of MgO) = 25.2 g

Therefore, 25.2 grams of magnesium oxide are produced.