Question

Сума порошкив алюмінію і міді 10 г обробили розчином сульфатної кислоти. пид час реакции видилився объемом 6,72 л ( н н.у). Обчислить масову частку металив у сумиши СРОЧНО​

Answer 1

Ответ:

w(Cu) = 46% , w(Al)= 54%

Объяснение:

   m                                             6.72

2Al + 3H2SO4 ⇒ Al2(SO4)3 + 3H2

54                                              67.2

Сu + H2SO4 ≠ H→Cu

m(Al) = 5.4 г.

m(Cu) = 10 - 5.4 = 4.6 г

w(Al) = 5.4/10 = 54%

w(Cu) = 100 - 54 = 46$

Answer 2

I can answer your question in Ukrainian. You asked me to answer in detail. The sum of aluminum and copper powders of 10 g was treated with a solution of sulfuric acid. During the reaction, a volume of 6.72 L (n. p.) was released. Calculate the mass fraction of metals in the mixture URGENTLY.

The reaction of aluminum with sulfuric acid can be written as:

$$ 2Al + 3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 3H_2 $$

The reaction of copper with sulfuric acid can be written as:

$$ Cu + H_2SO_4 \rightarrow CuSO_4 + H_2 $$

The total volume of hydrogen gas released is 6.72 L at normal conditions. Using the ideal gas law, we can find the number of moles of hydrogen gas:

$$ n = \frac{PV}{RT} = \frac{6.72 \times 10^{-3} \times 1.013 \times 10^5}{8.314 \times 273.15} \approx 0.030 \text{ mol} $$

Since one mole of hydrogen gas is produced for every mole of aluminum or copper reacted, the total number of moles of metals in the mixture is also 0.030 mol. The mass of aluminum and copper in the mixture can be found by multiplying the number of moles by their respective molar masses:

$$ m_{Al} = n_{Al} \times M_{Al} = x \times 0.030 \times 26.98 \approx 0.810x \text{ g} $$

$$ m_{Cu} = n_{Cu} \times M_{Cu} = (1-x) \times 0.030 \times 63.55 \approx 1.907(1-x) \text{ g} $$

where $x$ is the mole fraction of aluminum in the mixture. The mass of the mixture is given as 10 g, so we can write an equation:

$$ m_{Al} + m_{Cu} = 10 $$

Substituting the expressions for $m_{Al}$ and $m_{Cu}$, we get:

$$ 0.810x + 1.907(1-x) = 10 $$

Solving for $x$, we get:

$$ x = \frac{10 - 1.907}{0.810 - 1.907} \approx 0.685 $$

Therefore, the mole fraction of aluminum in the mixture is 0.685, and the mole fraction of copper is 1 - 0.685 = 0.315. The mass fraction of aluminum and copper can be found by dividing the mass by the total mass of the mixture:

$$ w_{Al} = \frac{m_{Al}}{10} = \frac{0.810x}{10} \approx 0.055 $$

$$ w_{Cu} = \frac{m_{Cu}}{10} = \frac{1.907(1-x)}{10} \approx 0.045 $$

Therefore, the mass fraction of aluminum in the mixture is 0.055, and the mass fraction of copper is 0.045. I hope this helps you with your problem.Ответь+подробно.++Сума+порошкив+алюмінію+і+міді+10+г+обробили+розчином+сульфатної+кислоти.+пид+час+реакции+видилився+объемом+6,72+л+(+н+н.у).+Обчислить+масову+частку+металив+у+сумиши+++СРОЧНО​+