Question

Вычислить объем кислорода, который потребуется для сжигания 11,2 литра бутана.

Answer 1

Вычислить объем кислорода который потребуется...

Answer 2

Calculating the Volume of Oxygen Required to Burn 11.2 Liters of Butane

To calculate the volume of oxygen required to burn 11.2 liters of butane, we can use the balanced chemical equation for the combustion of butane, which is:

2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

From this equation, we can see that 13 moles of oxygen are required to burn 2 moles of butane. Using the molar volume of gases at standard temperature and pressure (STP), which is 22.4 liters per mole, we can calculate the volume of oxygen required.

First, we need to convert the volume of butane from liters to moles, and then use the mole ratio from the balanced chemical equation to find the volume of oxygen required.

The molar volume of gases at STP is 22.4 liters per mole.

Calculation

Given: Volume of butane (C4H10) = 11.2 liters

1. Convert the volume of butane from liters to moles: - Molar volume of gases at STP = 22.4 liters/mole - Moles of butane = Volume of butane / Molar volume of gases at STP - Moles of butane = 11.2 liters / 22.4 liters/mole - Moles of butane = 0.5 moles

2. Use the mole ratio from the balanced chemical equation to find the volume of oxygen required: - From the balanced chemical equation: 2 moles of butane require