How many grams of chlorine gas must be reacted with excess sodium iodide if 10.0 grams of sodium chloride are needed?
2 Nal + Cl2= 2NaCl +I2
To solve this problem, we need to use the balanced chemical equation provided and apply the concept of stoichiometry.
Step 1: Find the molar mass of sodium chloride (NaCl)
The molar mass of sodium chloride (NaCl) is 58.44 g/mol. This is calculated as follows:
Na: 22.99 g/mol + Cl: 35.45 g/mol = 58.44 g/mol
Step 2: Determine the moles of sodium chloride (NaCl) required
To determine the moles of sodium chloride (NaCl) required, we can use the given mass and the molar mass calculated in Step 1.
Moles of NaCl = Mass of NaCl / Molar mass of NaCl
Moles of NaCl = 10.0 g / 58.44 g/mol = 0.171 mol
Step 3: Use stoichiometry to relate the moles of sodium chloride (NaCl) to chlorine gas (Cl2)
From the balanced chemical equation, we can see that 1 mole of chlorine gas (Cl2) reacts to produce 2 moles of sodium chloride (NaCl).
This means that the moles of Cl2 required will be half of the moles of NaCl.
Moles of Cl2 = 0.171 mol / 2 = 0.086 mol
Step 4: Calculate the grams of chlorine gas (Cl2) required
To calculate the grams of chlorine gas (Cl2) required, we use the molar mass of chlorine gas (Cl2), which is 70.90 g/mol.
Grams of Cl2 = Moles of Cl2 × Molar mass of Cl2
Grams of Cl2 = 0.086 mol × 70.90 g/mol = 6.10 grams
Therefore, 6.10 grams of chlorine gas must be reacted with excess sodium iodide to obtain 10.0 grams of sodium chloride.