Question

При окислении пропаналя аммиачным раствором оксида серебра образовалось 43,2 г. серебра. Какая масса пропаналя вступила в реакцию?

Answer 1

По реакции 2 из 0.4 моль серебра 0.2 моль пропаналя!
58 г х 0.2 моль=11.6 г ПРОПАНАЛЯ!!

Answer 2

To answer this question, we need to write the balanced chemical equation for the oxidation of propanal by ammoniacal silver oxide solution. The equation is:

$$CH_3CH_2CHO + 2Ag(NH_3)_2OH \rightarrow CH_3CH_2COOH + 2Ag + 4NH_3 + H_2O$$

According to the equation, one mole of propanal reacts with two moles of silver to produce two moles of silver. The molar mass of propanal is 58 g/mol and the molar mass of silver is 108 g/mol. Therefore, the mass of propanal that reacts with 43.2 g of silver can be calculated by using the following ratio:

$$\frac{58 \text{ g propanal}}{2 \times 108 \text{ g silver}} = \frac{x \text{ g propanal}}{43.2 \text{ g silver}}$$

Solving for x, we get:

$$x = \frac{58 \times 43.2}{2 \times 108}$$

$$x = 13.8 \text{ g propanal}$$

Therefore, the mass of propanal that entered the reaction is 13.8 g. You can find more information about this reaction and similar ones in the web search results . I hope this helps.