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soldier1979 [14.2K]
2 years ago
15

how much hydrogen will be released during the interaction of potassium weighing 8 grams with hydrochloric acid

Chemistry
1 answer:
krek1111 [17]2 years ago
4 0

Answer:

With an understanding of the ideal gas laws, it is now possible to apply these principles to chemical stoichiometry problems. For example, zinc metal and hydrochloric acid (hydrogen chloride dissolved in water) react to form zinc (II) chloride and hydrogen gas according to the equation shown below:

2 HCl (aq) + Zn (s) → ZnCl2 (aq) + H2 (g)

Explanation:

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\large\colorbox{lightblue}{Xxᴊᴀsʜ13xX}

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Answer:

Here's what I get  

Explanation:

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CH₂=CHCH₂CH₂CH₂CH₃ —  hex-1-ene is the preferred IUPAC name (PIN). 1-Hexene is accepted

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What poisonous gas is found in the exhaust fumes of car engines?
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The toxic gar expelled from the reaction between gasoline and oxygen in the vehicle's engine is both Carbon dioxide and monoxide
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Calculate how many valence electrons are in H2CO, N2, Cl2?
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3 years ago
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The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process.... In the first step, manganes
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Explanation :

First we have to calculate the mass of MnO_2.

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Let the mass of MnCO_3 be, 'x' grams.

From the balanced reaction, we conclude that

As, (2\times 115)g of MnCO_3 react to give (2\times 87)g of MnO_2

So, xg of MnCO_3 react to give \frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg of MnO_2

And as we are given that the yield produced from the first step is, 65 % that means,

60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg

The mass of MnO_2 obtained = 0.4542x g

Now we have to calculate the mass of Mn.

The second step balanced chemical reaction is:

3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3

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From the balanced reaction, we conclude that

As, (3\times 87)g of MnO_2 react to give (3\times 55)g of Mn

So, 0.4542xg of MnO_2 react to give \frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg of Mn

And as we are given that the yield produced from the second step is, 80 % that means,

80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg

The mass of Mn obtained = 0.2296x g

The given mass of Mn = 8.0 kg = 8000 g     (1 kg = 1000 g)

So, 0.2296x = 8000

x = 34843.20 g = 34.84 kg = 35 kg

Therefore, the mass of MnCO_3 required are, 35 kg

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Answer:

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