The equation for the synthesis of diborane, B2H6, a highly reactive boron hyride, is shown below. Calculate the mass of diborane produced from the reaction of 10.0 g of boron and 2.30 g of hydrogen.

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1. Write the balanced chemical equation for the synthesis of diborane: $2B + 3H_2 \rightarrow B_2H_6$

2. Calculate the molar masses of the reactants and the product:

   • Boron (B): $10.81 \, \text{g/mol}$
   • Hydrogen (H_2): $2 \times 1.01 \, \text{g/mol} = 2.02 \, \text{g/mol}$
   • Diborane (B_2H_6): $2 \times 10.81 \, \text{g/mol} + 6 \times 1.01 \, \text{g/mol} = 27.64 \, \text{g/mol}$

3. Convert the masses of boron and hydrogen to moles:

   • Moles of boron: $\frac{10.0 \, \text{g}}{10.81 \, \text{g/mol}} = 0.925 \, \text{mol}$
   • Moles of hydrogen: $\frac{2.30 \, \text{g}}{2.02 \, \text{g/mol}} = 1.139 \, \text{mol}$

4. Determine the limiting reactant:

   • According to the balanced equation, 2 moles of boron react with 3 moles of hydrogen.
   • Calculate the required moles of hydrogen for 0.925 moles of boron: $0.925 \, \text{mol B} \times \frac{3 \, \text{mol H}_2}{2 \, \text{mol B}} = 1.3875 \, \text{mol H}_2$
   • Compare the required moles of hydrogen with the available moles: $1.3875 \, \text{mol H}_2 > 1.139 \, \text{mol H}_2$
   • Hydrogen is the limiting reactant.

5. Calculate the moles of diborane produced:

   • According to the balanced equation, 3 moles of hydrogen produce 1 mole of diborane.
   • Moles of diborane produced: $1.139 \, \text{mol H}_2 \times \frac{1 \, \text{mol B}_2H_6}{3 \, \text{mol H}_2} = 0.3797 \, \text{mol B}_2H_6$

6. Convert the moles of diborane to mass:

   • Mass of diborane: $0.3797 \, \text{mol B}_2H_6 \times 27.64 \, \text{g/mol} = 10.49 \, \text{g}$

Therefore, the mass of diborane produced from the reaction is $10.49 \, \text{g}$.