Elements A and B react according to the following equation. 4A + B2 →2A2B The molar mass of A is 12 g/mol. The molar mass of B is 4 g/mol. The molar mass of A2B is 28 g/mol. When the goal of the reaction is to produce 112 g of the product, how many grams of reactant A should be present initially?
To find out how many grams of reactant A should be present initially, we need to calculate the stoichiometry of the reaction and determine the amount of A needed to produce 112 g of A2B.
From the balanced equation, we can see that 4 moles of A react with 1 mole of B2 to produce 2 moles of A2B. This means that the molar ratio of A to A2B is 4:2 or 2:1.
The molar mass of A is given as 12 g/mol. Therefore, 1 mole of A would have a mass of 12 g.
Since we need to produce 112 g of A2B, we first calculate the number of moles of A2B present in 112 g:
Moles of A2B = Mass of A2B / Molar mass of A2B
Moles of A2B = 112 g / 28 g/mol
Moles of A2B = 4 moles
According to the stoichiometry of the reaction, we know that for every 2 moles of A2B produced, we need 4 moles of A. Therefore, if we need 4 moles of A2B, we would need 8 moles of A.
Now, we calculate the mass of A required to produce 8 moles:
Mass of A = Moles of A x Molar mass of A
Mass of A = 8 moles x 12 g/mol
Mass of A = 96 g
Therefore, to produce 112 g of A2B, we would need 96 g of reactant A present initially.