Question

Starting with the following equation, bcl₃(g) + lialh‚„(s) → b₂h₄(g) + lialcl‚„(s), calculate the moles of bcl₃ that will be required to produce 735 grams of b₂h₄?

Answer 1

To calculate the moles of BCl3 required to produce 735 grams of B2H4, use the balanced chemical equation and the molar masses of the substances involved.

Step-by-step explanation:

To calculate the moles of BCl3 required to produce 735 grams of B2H4, we need to use the balanced chemical equation and the molar masses of the substances involved.

The balanced chemical equation is:

BCl3(g) + LiAlH4(s) → B2H4(g) + LiAlCl4(s)

From the equation, we can see that 1 mole of BCl3 reacts to produce 1 mole of B2H4. Therefore, the moles of BCl3 required can be calculated as follows:

735 g B2H4 * (1 mole BCl3 / molar mass of B2H4) = moles of BCl3

First, let's calculate the molar mass of B2H4. The molar mass of B is approximately 10.81 g/mol, and the molar mass of H is approximately 1.008 g/mol. So, the molar mass of B2H4 is:

Molar mass of B2H4 = (2 * molar mass of B) + (4 * molar mass of H)

Molar mass of B2H4 = (2 * 10.81 g/mol) + (4 * 1.008 g/mol)

Molar mass of B2H4 = 21.62 g/mol + 4.032 g/mol

Molar mass of B2H4 = 25.652 g/mol

Now, let's substitute the values in to calculate the moles of BCl3:

735 g B2H4 * (1 mole BCl3 / 25.652 g B2H4) = moles of BCl3

Calculate the result to find the moles of BCl3.