Passage VI

When hydrogen and oxygen are mixed under certain conditions, H2O will form. Students did the following experiments to study how H2O forms.

Experiment 1

A thick-walled gas syringe fitted with a sparking device was filled with 20 mL of hydrogen gas (H2) and 10 mL of oxygen gas (O2)，as shown in Figure 1.

The syringe plunger was then locked in place and the gases were sparked. A reaction took place, forming droplets of water. The plunger was then released. The gas in the syringe was allowed to adjust to room temperature and pressure, and the final volume of gas was recorded. The remaining gas, if any, was then analyzed to determine its composition. The procedure was repeated with different volumes of gases, and the results were recorded in Table 1.

Since it is known that equal numbers of molecules of different gases occupy equal volumes at the same pressure and temperature, the following equation was suggested:

2 H2 + O2 → H2O

Experiment 2

In the apparatus shown in Figure 2, a stream of H2 was passed over hot copper oxide (CuO), producing H2O vapor. The vapor was then absorbed by calcium chloride (CaCl2).

The changes in the masses of the tube sections containing the CuO and CaCl2 were used to calculate the mass of CuO reacted and the mass of H2O formed. It was determined that 1 molecule of H2O was produced for every CuO reacted:

CuO + H2 → Cu + H2O

35. Which of the following, if it had occurred, would probably NOT have caused an error in interpreting the results of Experiment 2 ?

Answer and Explanation

Your Answer is

Correct Answer is A

Explanation

Look at the second line of the second paragraph of Experiment 2...were used to calculate the mass of CuO reacted and the mass of H2O formed. It can be known that the purpose of the experiment is to measure the participating reactions The mass of CuO, and the mass of H2O produced by the reaction.

If the situation mentioned in option B occurs, then the quality of CuO that is really participating in the reaction cannot be measured, so it is excluded;

If the situation mentioned in option C occurs , then the quality of the product H2O cannot be measured, so it is excluded;

If the situation mentioned in option D occurs, CuO and H2 The reaction also generates some other products, so it is impossible to judge how much CuO participated in the reaction of generating H2O, and if other products are liquid, it may also lead to the inaccurate measurement of H 2The quality of O, so it is also excluded;

Only the H2 of option A contains impurities that do not participate in the reaction, which will not affect the measurement results< /p>