Based solely on the amount of available carbon, how many grams of sodium oxalate, Na2C2O4, could be obtained from 175 g of C6H6? (Assume that no loss of carbon occurs in any of the reactions needed to produce the Na2C2O4.)
To find the amount of sodium oxalate that can be obtained from 175 g of benzene (C6H6), we need to calculate the amount of carbon present in benzene and then determine the stoichiometry between carbon and sodium oxalate (Na2C2O4) to find the mass of sodium oxalate.
1. Step 1: Calculate the molar mass of benzene (C6H6).
The molar mass of carbon (C) is 12.01 g/mol, and the molar mass of hydrogen (H) is 1.01 g/mol. Benzene (C6H6) consists of 6 carbon atoms and 6 hydrogen atoms.
Molar mass of C6H6 = (6 * molar mass of C) + (6 * molar mass of H) = (6 * 12.01 g/mol) + (6 * 1.01 g/mol) = 78.11 g/mol
2. Step 2: Calculate the moles of benzene (C6H6).
To find the number of moles of benzene, divide the given mass (175 g) by the molar mass of C6H6.
moles of C6H6 = Mass of C6H6 / molar mass of C6H6 = 175 g / 78.11 g/mol ≈ 2.242 mol
3. Step 3: Determine the stoichiometry between carbon (C) and sodium oxalate (Na2C2O4).
The balanced chemical equation for the formation of sodium oxalate (Na2C2O4) from carbon (C) is as follows:
2 C + Na2CO3 + H2O -> 2 Na2C2O4 + H2
From this equation, we can see that 2 moles of carbon (C) react to produce 2 moles of sodium oxalate (Na2C2O4).
4. Step 4: Calculate the moles of sodium oxalate (Na2C2O4) produced.
As we determined in Step 3, the stoichiometry between carbon and sodium oxalate is 1:1. Therefore, the moles of sodium oxalate produced will be equal to the number of moles of carbon present in benzene.
moles of Na2C2O4 = moles of C6H6 ≈ 2.242 mol
5. Step 5: Calculate the mass of sodium oxalate (Na2C2O4) produced.
To find the mass of sodium oxalate, multiply the moles of sodium oxalate by its molar mass.
Mass of Na2C2O4 = moles of Na2C2O4 * molar mass of Na2C2O4
The molar mass of Na2C2O4 = (2 * molar mass of Na) + (2 * molar mass of C) + (4 * molar mass of O)
= (2 * 22.99 g/mol) + (2 * 12.01 g/mol) + (4 * 16.00 g/mol)
= 133.99 g/mol
Substituting the values:
Mass of Na2C2O4 = 2.242 mol * 133.99 g/mol ≈ 300.7 g
Therefore, approximately 300.7 grams of sodium oxalate (Na2C2O4) can be obtained from 175 g of benzene (C6H6).