Elements A and B react according to the following balanced equation. 3A2+2B → 2A3B3A2+2B → 2A3B The molar mass of element A is 4g/mol. The molar mass of element B is 16g/mol. When the initial mass of element A is 48g, which mass of element B should be present? (1 point) Responses 192g 192g 128g 128g 96g 96g 64g
The molar ratio between A and B in the balanced equation is 3:2. This means that for every 3 moles of A that react, 2 moles of B are required.
Given that the initial mass of A is 48g and the molar mass of A is 4g/mol, we can calculate the number of moles of A present:
48g / 4g/mol = 12 moles of A
Using the molar ratio, we can determine the number of moles of B required:
12 moles of A * (2 moles of B / 3 moles of A) = 8 moles of B
Now we can calculate the mass of B that corresponds to 8 moles:
8 moles of B * 16g/mol = 128g
Therefore, 128g of element B should be present. The correct answer is 128g.