Question

Elements A and B react according to the following balanced equation.

3A2+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

64g
64g

128g
128g

96g
96g

192g

Answer 1

To determine the mass of element B needed for the reaction, we start with the balanced chemical equation:

\[ 3A_2 + 2B \rightarrow 2A_3B \]

1. **Calculate the moles of A**:
- The molar mass of element A is 4 g/mol.
- The initial mass of A is 48 g.

\[
\text{Moles of A} = \frac{\text{mass}}{\text{molar mass}} = \frac{48 \, \text{g}}{4 \, \text{g/mol}} = 12 \, \text{mol}
\]

2. **Determine how many moles of \( A_2 \) are present**:
- Since \( A_2 \) consists of 2 moles of A, we need to calculate the moles of \( A_2 \):

\[
\text{Moles of } A_2 = \frac{12 \, \text{mol}}{2} = 6 \, \text{mol}
\]

3. **Using the stoichiometry of the balanced equation**:
- From the equation, 3 moles of \( A_2 \) react with 2 moles of B.
- Thus, for 6 moles of \( A_2 \), the amount of B required can be calculated:

\[
\text{Moles of B} = \left( \frac{2 \, \text{moles of B}}{3 \, \text{moles of } A_2} \right) \times 6 \, \text{moles of } A_2 = 4 \, \text{moles of B}
\]

4. **Calculate the mass of B**:
- The molar mass of element B is 16 g/mol.

\[
\text{Mass of B} = \text{moles} \times \text{molar mass} = 4 \, \text{mol} \times 16 \, \text{g/mol} = 64 \, \text{g}
\]

Therefore, the mass of element B that should be present is **64 g**.

So the answer is: **64g**.