To solve this problem, we need to use stoichiometry, which is a method to calculate the amounts of reactants and products in a chemical reaction based on the balanced equation.
First, we need to calculate the number of moles of element A present in 48 grams:
48 g / 4 g/mol = 12 mol A
Next, we use the stoichiometry of the balanced equation to find the number of moles of element B required to react with 12 mol A:
3 mol A2 / 2 mol B = 2 mol A3B / 1 mol B
12 mol A / 3 mol A2 * 2 mol A3B / 1 mol B = 8 mol B
Finally, we convert the number of moles of element B to mass using its molar mass:
8 mol B * 16 g/mol = 128 grams
Therefore, the answer is (a) 128 grams of element B should be present.
Elements A and B react according to the following balanced equation.
3A2 + 2B → 2A3B
The molar mass of element A is 4 g/mol. The molar mass of element B is 16 g/mol. When the initial mass of element A is 48 grams, which mass of element B should be present?
Responses-
a- 128 grams
b- 192 grams
c- 64 grams
d- 96 grams
1 answer
1 answer