Final answer:
a. 132.01 g of Na2O are produced when 57.5 g of Na reacts. b. 8.00 g of O2 are required for reaction with 18.0 g of Na. c. 0.41 g of O2 is needed to produce 75.0 g of Na2O.
Step-by-step explanation:
a. To determine the number of grams of Na2O produced, we need to use the molar mass of Na2O. Given that 1 mole of Na2O has a molar mass of 77.96 g/mol, we can set up a proportion:
4 Na = 57.5 g Na => 1 mole Na = x moles Na => 77.96 g Na2O = y grams Na2O
Solving this proportion, we find that 57.5 g of Na reacts to form (77.96 g/mol) / (46.0 g/mol) = 1.69 moles of Na2O. Since 1 mole of Na2O has a molar mass of 77.96 g/mol, 1.69 moles of Na2O would have a mass of 1.69 moles x 77.96 g/mol = 132.01 g Na2O.
b. To determine the mass of O2 required for the reaction, we can use the molar ratio between Na and O2. From the balanced equation, we see that 4 moles of Na react with 1 mole of O2. Given that 1 mole of O2 has a molar mass of 32.00 g/mol, we can set up the following proportion:
4 Na = 22.99 g Na => 1 mole Na = x moles Na => 1 mole O2 = y moles O2 => 32.00 g O2 = z grams O2
Solving this proportion, we find that 18.0 g of Na would require (1 mole O2) / (4 moles Na) = 0.25 moles of O2. Since 1 mole of O2 has a molar mass of 32.00 g/mol, 0.25 moles of O2 would have a mass of 0.25 moles x 32.00 g/mol = 8.00 g O2.
c. To determine the mass of O2 needed to produce 75.0 g of Na2O, we can use a similar approach as in part b. From the balanced equation, we see that for every mole of Na2O produced, 1 mole of O2 is required. Given that 1 mole of O2 has a molar mass of 32.00 g/mol, we can set up the following proportion:
4 Na = 22.99 g Na => 1 mole Na = x moles Na => 1 mole O2 = y moles O2 => 32.00 g O2 = z grams O2
Solving this proportion, we find that 75.0 g of Na2O would require (1 mole O2) / (77.96 g/mol) = 0.0128 moles of O2. Since 1 mole of O2 has a molar mass of 32.00 g/mol, 0.0128 moles of O2 would have a mass of 0.0128 moles x 32.00 g/mol = 0.41 g O2.