Question

Объём н.у и состав газа выделившегося при взаимодействии 6,4 г меди с концентрированной серной кислотой

Answer 1

Дано:
m(Cu)=6.4г
Найти:
V(SO2)-?
Решение:
6.4г. Xл
Cu + 2H2SO4 = SO2 + 2H2O + CuSO4
64г. 22.4л
X=6.4г×22.4л:64г=2.24л
Ответ: выделился оксид серы (IV) массой 2.24л

Answer 2

Reaction Details

The reaction between 6.4 g of copper and concentrated sulfuric acid can be represented by the following chemical equation: Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O.

Volume of Gas Evolved

The volume of gas evolved can be calculated using the ideal gas law, which states that PV = nRT, where: - P is the pressure of the gas, - V is the volume of the gas, - n is the number of moles of the gas, - R is the ideal gas constant, and - T is the temperature in Kelvin.

Given the reaction, the molar ratio between copper and sulfur dioxide is 1:1. Therefore, the number of moles of sulfur dioxide produced is equal to the number of moles of copper used in the reaction.

Calculating the Number of Moles

The molar mass of copper (Cu) is approximately 63.55 g/mol Using this information, we can calculate the number of moles of copper involved in the reaction.

Number of moles of Cu = Mass of Cu / Molar mass of Cu Number of moles of Cu = 6.4 g / 63.55 g/mol

Molar Volume of Gas at STP

The molar volume of a gas at standard temperature and pressure (STP) is approximately 22.4 L/mol.

Calculating the Volume of Sulfur Dioxide

Using the molar volume of gas at STP, we can calculate the volume of sulfur dioxide produced.

Volume of SO2 = Number of moles of Cu * Molar volume of gas at STP Volume of SO2 = (6.4 g / 63.55 g/mol) * 22.4 L/mol

Result

By performing the calculations, we can determine the volume of sulfur dioxide gas produced when 6.4 g of copper reacts with concentrated sulfuric acid.