Small amounts of chlorine gas can be gener- ated in the laboratory from the reaction of manganese(IV) oxide with hydrochloric acid: 4 HCl(aq) + MnO2(s) →
2 H2O(l) + MnCl2(s) + Cl2(g). What mass of Cl2 can be produced from 41.8 g
of MnO2 with an excess of HCl(aq)?
HCl(aq) + MnO2(s) →
2 H2O(l) + MnCl2(s) + Cl2(g).
mols MnO2 = grams/molar mass = ?
mols Cl2 produced = mols Cl2 produced since the reaction of 1 mol MnO2 produces 1 mol Cl2.
Then grams Cl2 = mols Cl2 x molar mass Cl2 = ?
To calculate the mass of Cl2 produced from the reaction, we need to use the stoichiometry of the balanced chemical equation. Here's how you can find the answer:
Step 1: Write and balance the chemical equation:
4 HCl(aq) + MnO2(s) → 2 H2O(l) + MnCl2(s) + Cl2(g)
Step 2: Find the molar mass of MnO2 and Cl2:
The molar mass of MnO2 is 86.94 g/mol (1 Mn atom + 2 O atoms).
The molar mass of Cl2 is 70.90 g/mol (2 Cl atoms).
Step 3: Convert the given mass of MnO2 to moles:
Using the molar mass of MnO2, we can convert grams to moles:
41.8 g MnO2 * (1 mol MnO2 / 86.94 g MnO2) = 0.480 mol MnO2
Step 4: Determine the stoichiometric ratio of MnO2 to Cl2:
From the balanced chemical equation, we can see that 1 mol of MnO2 produces 1/2 mol of Cl2.
Step 5: Use stoichiometry to calculate the moles of Cl2 produced:
0.480 mol MnO2 * (1/2 mol Cl2 / 1 mol MnO2) = 0.240 mol Cl2
Step 6: Convert moles of Cl2 to grams:
Using the molar mass of Cl2, we can convert moles to grams:
0.240 mol Cl2 * (70.90 g Cl2 / 1 mol Cl2) = 16.936 g Cl2
Therefore, 41.8 g of MnO2 can produce 16.936 g of Cl2.