Определить объем ацетилена, кот. образуется при взаимодействии 6,4 г. карбида кальция и 1,8 г. воды, если выход 90%.
To determine the volume of acetylene formed from 6.4 grams of calcium carbide and 1.8 grams of water with a 90% yield, we first need to calculate the limiting reagent. In this case, it is water because it is present in the smallest amount.Next, we need to convert the mass of water to moles by dividing it by the molar mass of water (18 g/mol). This gives us 0.1 moles of water.
Using the balanced chemical equation for the reaction between calcium carbide and water, we can see that 1 mole of calcium carbide produces 1 mole of acetylene: CaC2 + 2H2O → C2H2 + Ca(OH)2
Therefore, 0.1 moles of water will produce 0.1 moles of acetylene.
To convert moles to volume, we need to use the ideal gas law, PV = nRT. Assuming standard temperature and pressure (STP), the volume of 0.1 moles of acetylene gas would be 2.24 liters (0.1 mol x 22.4 L/mol).
Therefore, the volume of acetylene formed is 2.24 liters with a 90% yield from 6.4 grams of calcium carbide and 1.8 grams of water. Keep in mind that this is the theoretical maximum volume and the actual volume may be slightly less due to other factors such as the reaction not going to completion or the formation of other by-products.
Just remember, chemistry is all about balancing equations and calculating everything down to the last mole.
