C6H6(l) + Cl2(g) --> C6Cl6(s) + HCl(g)
Consider the unbalanced equation above. What is the maximum mass of each of the products that can be formed when 72.5 g of C6H6 and 49.0 g Cl2 react?
1. Balance the equation.
2a Convert 72.5 g C6H6 to moles. mol = g/molar mass
2b. Do the same for 49.0g Cl2.
3a. Using the coefficients in the balanced equation, convert moles C6H6 to moles of the product.
3b. Do the same for Cl2.
3c. You will get two different answers; obviously both can't be right. The correct answer to choose in limiting reagent problems is ALWAYS the smaller value and the reagent providing that number is the limiting reagent.
4. Using the smaller number of mols, convert to grams. g = mols x molar mass.
Would 20.0g of steam at 100 deg centi be enough to melt 20.0 g of ice at 0 degre cent
To find the maximum mass of each product that can be formed, we need to balance the equation first.
Let's balance the equation step by step:
C6H6(l) + Cl2(g) → C6Cl6(s) + HCl(g)
Count the number of carbon (C) atoms on each side of the equation:
On the left-hand side (LHS), there are 6 carbon atoms in C6H6.
On the right-hand side (RHS), there are 6 carbon atoms in C6Cl6.
Count the number of hydrogen (H) atoms on each side of the equation:
On the LHS, there are 6 hydrogen atoms in C6H6.
On the RHS, there is 1 hydrogen atom in HCl.
Count the number of chlorine (Cl) atoms on each side of the equation:
On the LHS, there are 2 chlorine atoms in Cl2.
On the RHS, there are 6 chlorine atoms in C6Cl6 and 1 chlorine atom in HCl.
To balance the equation, we need to place coefficients in front of the chemical formulas to make the number of atoms the same on both sides of the equation:
C6H6(l) + 3Cl2(g) → C6Cl6(s) + 6HCl(g)
Now that the equation is balanced, we can use stoichiometry to find the maximum mass of each product that can be formed.
First, we need to calculate the number of moles of C6H6 and Cl2 using their molar masses:
Molar mass of C6H6: 12.01 g/mol (carbon) + 1.01 g/mol (hydrogen) = 78.11 g/mol
Molar mass of Cl2: 35.45 g/mol (chlorine) + 35.45 g/mol (chlorine) = 70.90 g/mol
Number of moles of C6H6 = mass of C6H6 / molar mass of C6H6 = 72.5 g / 78.11 g/mol = 0.9288 mol
Number of moles of Cl2 = mass of Cl2 / molar mass of Cl2 = 49.0 g / 70.90 g/mol = 0.6911 mol
Based on the balanced equation, the molar ratio between C6H6 and C6Cl6 is 1:1. This means that for every 1 mole of C6H6, 1 mole of C6Cl6 is produced.
Maximum moles of C6Cl6 = number of moles of C6H6 = 0.9288 mol
To calculate the mass of C6Cl6, we multiply the maximum moles of C6Cl6 by its molar mass:
Mass of C6Cl6 = maximum moles of C6Cl6 × molar mass of C6Cl6
= 0.9288 mol × (6 × 12.01 g/mol + 6 × 35.45 g/mol)
= 0.9288 mol × (72.06 g/mol)
= 67.02 g
Therefore, the maximum mass of C6Cl6 that can be formed is 67.02 grams.
Similarly, using the balanced equation, the molar ratio between C6H6 and HCl is 1:6. This means that for every 1 mole of C6H6, 6 moles of HCl are produced.
Maximum moles of HCl = number of moles of C6H6 × 6 = 0.9288 mol × 6 = 5.5728 mol
To calculate the mass of HCl, we multiply the maximum moles of HCl by its molar mass:
Mass of HCl = maximum moles of HCl × molar mass of HCl
= 5.5728 mol × (1 × 1.01 g/mol + 1 × 35.45 g/mol)
= 5.5728 mol × (36.46 g/mol)
= 202.682 g
Therefore, the maximum mass of HCl that can be formed is 202.682 grams.