10. Diborane, B_(2)H_(6) was used as rocket fuel previously. It reacts according to the following equation: B_(2)H_(6)+3O_(2)arrow B_(2)O_(3)+3H_(2)O How many moles of oxygen would be needed to fully combust 40 Liters of B_(2)H_(6) at STP?

To solve this problem, we need to use the balanced chemical equation and the given information to determine the number of moles of oxygen required to fully combust 40 liters of diborane ($B_{2}H_{6}$) at standard temperature and pressure (STP).<br /><br />Given information:<br />- Balanced chemical equation: $B_{2}H_{6} + 3O_{2} \rightarrow B_{2}O_{3} + 3H_{2}O$<br />- Volume of diborane: 40 liters<br /><br />Step 1: Calculate the number of moles of diborane ($B_{2}H_{6}$) using the ideal gas law.<br />At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 liters.<br />Number of moles of diborane = Volume of diborane / Volume of 1 mole of gas at STP<br />Number of moles of diborane = 40 liters / 22.4 liters/mole = 1.7857 moles<br /><br />Step 2: Use the balanced chemical equation to determine the number of moles of oxygen required.<br />According to the balanced equation, 1 mole of diborane ($B_{2}H_{6}$) requires 3 moles of oxygen ($O_{2}$) to fully combust.<br />Number of moles of oxygen required = 3 × Number of moles of diborane<br />Number of moles of oxygen required = 3 × 1.7857 moles = 5.3561 moles<br /><br />Therefore, 5.3561 moles of oxygen would be needed to fully combust 40 liters of diborane ($B_{2}H_{6}$) at STP.