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
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.
I'll have to do some math to answer this question, but I promise not to make it too molar-dramatic! Let's break it down. We know that the molar mass of element A is 4 g/mol, so if we have 48 grams of A, we have 48/4 = 12 moles of A. According to the balanced equation, 3 moles of A2 reacts with 2 moles of B to form 2 moles of A3B.
Now, let's do some mole-ath! Since 3 moles of A2 react with 2 moles of B, we have a ratio of 3:2. If we have 12 moles of A2, we can set up a proportion to find out how many moles of B we need.
(3 moles A2 / 2 moles B) = (12 moles A2 / x moles B)
Cross-multiplying, we get 3x = 24, and solving for x, we find x = 24/3 = 8 moles of B.
Now, we know that the molar mass of element B is 16 g/mol, so if we have 8 moles of B, we have 8 * 16 = 128 grams of B.
So, the answer is (a) 128 grams. Tadaa!
To determine the mass of element B, we can use the stoichiometry of the balanced equation. According to the equation, the ratio of A2 to B is 3:2.
Given that the molar mass of element A is 4 g/mol, we can calculate the number of moles of A2 present:
Number of moles of A2 = mass of A2 / molar mass of A2 = 48 g / 4 g/mol = 12 mol
From the stoichiometry of the equation, we know that 3 moles of A2 react with 2 moles of B. To find the number of moles of B, we can set up a proportion:
(2 mol B / 3 mol A2) = (x mol B / 12 mol A2)
Simplifying the proportion, we have:
2/3 = x/12
Cross multiplying:
3x = 24
Solving for x:
x = 24/3 = 8
Therefore, the number of moles of B required is 8 mol.
Next, we can calculate the mass of element B:
Mass of B = number of moles of B × molar mass of B
= 8 mol × 16 g/mol
= 128 g
Therefore, the mass of element B that should be present is 128 grams.
Answer: a- 128 grams
To solve this problem, first, we need to calculate the number of moles of element A present in the initial mass of 48 grams. We can use the molar mass of element A (4 g/mol) to do this.
Number of moles of A = Mass of A / Molar mass of A
Number of moles of A = 48 grams / 4 g/mol
Number of moles of A = 12 moles
Next, using the balanced equation, we can relate the stoichiometry of element A and element B. The ratio of moles of A to moles of B is 3:2.
According to the equation: 3A2 + 2B → 2A3B
Moles of A / 3 = Moles of B / 2
Substituting the value for moles of A (12 moles) into the equation, we can solve for moles of B:
12 moles / 3 = Moles of B / 2
Moles of B = 24 moles
Finally, we can calculate the mass of element B by multiplying the number of moles by the molar mass of B (16 g/mol):
Mass of B = Moles of B * Molar mass of B
Mass of B = 24 moles * 16 g/mol
Mass of B = 384 grams
Therefore, the mass of element B that should be present is 384 grams. However, none of the given options match this answer.