Answer:
1) Therefore, 8.79 * 10^23 oxygen molecules are needed to react completely with 135 grams of sodium.
2)1.24 x 10^3 g / (2 moles AlCl3/2 moles Al) = 622 grams of aluminum
Step-by-step explanation:
1)To determine the number of oxygen molecules needed to react completely with 135 grams of sodium, we need to use the balanced chemical equation for the reaction. The equation states that 4 moles of sodium (Na) react with 1 mole of oxygen (O2) to form 2 moles of sodium oxide (Na2O).
Given that 135 grams of sodium is equal to 135/23 = 5.87 moles, then 5.87 moles of sodium will react with 5.87 moles / 4 moles of sodium = 1.47 moles of oxygen.
So, 1.47 moles of oxygen is equal to 1.47 * Avogadro's number (6.022 * 10^23) = 8.79 * 10^23 oxygen molecules.
Therefore, 8.79 * 10^23 oxygen molecules are needed to react completely with 135 grams of sodium.
2)To determine the amount of aluminum needed to produce 7.5 x 10^30 molecules of aluminum chloride, we need to first determine the formula weight of aluminum chloride. The formula weight is calculated by summing up the atomic weights of all the elements in the formula:
Aluminum chloride: Al + Cl + Cl = 27 g/mol + 35.5 g/mol + 35.5 g/mol = 98 g/mol
Next, we need to convert the number of molecules of aluminum chloride to grams:
7.5 x 10^30 molecules x 98 g/mol/6.022 x 10^23 molecules = 1.24 x 10^3 grams
Since two moles of aluminum reacts with three moles of chlorine to produce two moles of aluminum chloride, the amount of aluminum needed can be calculated as follows:
1.24 x 10^3 g / (2 moles AlCl3/2 moles Al) = 622 grams of aluminum