Question

Який об'єм водню (н.у.) виділиться при взаємодії 0,2 моль алюмінію із хлоридною кислотою СРОЧНО​

Answer 1

2Al + 6HCl = 2AlCl3 + 3H2

μ(Al) = 0,2 моль

V(H2) - ?
Решение:

1) μ(H2) = 0,2 моль / 2 * 3 = 0,3 моль

2) V(H2) = 0,3 моль * 22,4 л/моль = 6,72 л

Ответ: 6,72 л

Answer 2

Ось розв‘язання цієї задачі :

Answer 3

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To answer your question, the volume of hydrogen (n.t.p.) that will be released during the interaction of 0.2 mol of aluminum with hydrochloric acid is calculated by the following formula:

$$V = nRT/P$$

where $n$ is the number of moles of hydrogen, $R$ is the universal gas constant, $T$ is the temperature in Kelvin, and $P$ is the pressure in atmospheres.

The balanced chemical equation for the reaction is:

$$2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2$$

From the equation, we can see that 0.2 mol of aluminum will produce 0.3 mol of hydrogen. Assuming that the temperature is 273 K and the pressure is 1 atm, we can substitute the values into the formula and get:

$$V = 0.3 \times 0.0821 \times 273 / 1$$

$$V = 6.72 L$$

Therefore, the volume of hydrogen (n.t.p.) that will be released is 6.72 liters. I hope this helps.Ответь+подробно.++Який+об\'єм+водню+(н.у.)+виділиться+при+взаємодії+0,2+моль+алюмінію+із+хлоридною+кислотою+СРОЧНО​+