Question

Сколько литров водорода (н.у.) образуется при взаимодействии достаточного количества магния с 200г 2,45%-го ратвора серной кислоты.

Answer 1

M(H2SO4)=w*m/100%                                                                                           m(H2SO4)=4.9 г                                                                                                      ню(H2SO4)=m(H2SO4)/M                                                                   ню(H2SO4)=0,05 моль                                                                                          ню(H2)=0.05 моль                                                                                                V(H2)=ню*V                                                                                                          V(H2)=0.1 г                                                                                                            Ответ: 0.1 г

Answer 2

Calculation of Hydrogen Gas Produced

To calculate the amount of hydrogen gas produced when a sufficient amount of magnesium reacts with a 200g 2.45% sulfuric acid solution, we need to use the balanced chemical equation for the reaction between magnesium and sulfuric acid.

The balanced chemical equation for the reaction is as follows:

Mg + H2SO4 → MgSO4 + H2

From the equation, we can see that one mole of magnesium reacts with one mole of sulfuric acid to produce one mole of hydrogen gas.

To calculate the moles of magnesium in the reaction, we need to convert the mass of magnesium to moles using its molar mass. The molar mass of magnesium (Mg) is approximately 24.31 g/mol.

Given that the mass of magnesium is not provided in the question, we cannot proceed with the calculation without this information. Please provide the mass of magnesium involved in the reaction so that we can calculate the amount of hydrogen gas produced.