Hydrogen gas can be prepared by reaction of zinc metal
with aqueo...

Question

Hydrogen gas can be prepared by reaction of zinc metal
with aqueous HCl:
Zn1s2 + 2 HCl1aq2¡ZnCl21aq2 + H21g2
(a) How many liters of H2 would be formed at 742 mm Hg and 15 °C if 25.5 g of zinc was allowed to react?

Accepted Answer

Hey everyone. So we have the following reaction that shows the decomposition of ammonium carbonate and at 28°C and 1.24 atmosphere test to determine the total volume of gas form 15.26 g of ammonium carbonate. That's completely decomposed total total number of malls. He goes to molds of NH three, that's one more Of c. 0. 2. That's one more Of H 20. And this will give us a total of four moles of gas. So now we need to convert from grams of ammonium carbonate, two moles of gas. We have 15.26 grams of ammonia carbonate. And in one mole that the Mueller mass, it's gonna be too. I'm 14.007 g plus eight Times 1.008g as well. 011 g plus three Times 15.999 gramps. Give us 96.086 g. And in one mole of ammonia carbonate, They're four moles of gas. This will give us 0.63 53 balls of gas. So now we can use the ideal gas out to find the total volume. It's going to be PB It goes N R. T. The pressure is 1.24 atmosphere. The volume is what we're looking for Number of moles. 0.6353. Our constant is 0.08 21. There's some atmosphere about about malls. I'm Calvin And the temperature is 28 degrees Celsius. We need to convert our temperature into Calvin 28°C. We need to add 273.15. This will give us 301 15 kelvin. So now we can put the values into the equation and solve for the volume. It's going to give us 1.24 atmosphere terms of volume Equals 0.63 malls From 0.08 later. Sounds atmosphere. What about moles and kelvin? I'm staying in one 15 kelvin. We get 1.24 times of volume. It was 15.707. So if we divide both sides by 1.24, We get a volume of 12.7 L. Thanks for watching my video and I hope it was helpful.

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