Question

Сколько литров кислорода (н. у.) расходуется при сжигание алюминия, массой 9 г.

Answer 1

9г       xл
4Al  +  3O2=2Al2O3
4*27  3*22,4
108г    67,2 л

9г--------------хл
108г--------67,2л
з=9*67,2/108 =5,6литра

Answer 2

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To answer your question, I will use the following chemical equation for the combustion of aluminum:

$$4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3$$

According to the equation, four moles of aluminum react with three moles of oxygen to produce two moles of aluminum oxide. To find the amount of oxygen consumed, we need to use the molar mass of aluminum and the molar volume of oxygen at standard conditions.

The molar mass of aluminum is 26.98 g/mol, so 9 g of aluminum corresponds to $\frac{9}{26.98} \approx 0.333$ moles of aluminum. Using the mole ratio from the equation, we can calculate the moles of oxygen needed for the reaction:

$$0.333 \text{ mol Al} \times \frac{3 \text{ mol O}_2}{4 \text{ mol Al}} = 0.25 \text{ mol O}_2$$

The molar volume of oxygen at standard conditions is 22.4 L/mol, so 0.25 moles of oxygen corresponds to $0.25 \times 22.4 \approx 5.6$ liters of oxygen. Therefore, the answer is:

5.6 liters of oxygen are consumed when 9 g of aluminum are burned.Ответь+подробно.+Сколько+литров+кислорода+(н.+у.)+расходуется+при+сжигание+алюминия,+массой+9+г.