Calculate the STP molar volume for hydrogen?
A 0.215g sample of zinc metal reacted with hydrochloric acid according to the following balanced equation : ZN(s) + 2 HCL(aq)----> ZNCL^2(aq) + H^2(g) The volume of hydrogen collected over water was 80.0mL at 20*c and a barometer reading of 756 mmHg. Calculate the STP molar volume for hydrogen.
Molar Volume of Hydrogen?
We did the typical magnesium in HCl experiment in school last week. For the post-lab questions its asks for the "literature" value of the molar volume of hydrogen. Its asks us to find the volume in the book (its not in there!). Does anyone know where I can find this, how to do this, or what it is? I got 8.3L/mol using my "experimental data." Not sure if thats correct. thanks!
Calculate the moles of Hydrogen gas produced and molar volume of Hydrogen gas at STP.?
The initial problem was: A 0.0795-g sample of Mg reacts with Hydrochloric acid to give 88.5 mL of Hydrogen gas at 25°C and 766 mm Hg. Show the calculation for the volume of Hydrogen gas at STP. Which came out to be 24.02 L/mol, but how do I calculate the moles of the Hydrogen gas produced and then find the molar volume of Hydrogen gas at STP?
Molar Volume of Hydrogen at STP?
The molar volume of gases at STP is always 22.4 L V = nRT / V at STP pressure = 1 atm and T = 273 K V = 1 mol x 0.0821 x 273 / 1 atm = 22.4 L
How much volume at STP does 1g of hydrogen occupy?
In many textbooks you must have read the statement,“One mole of any ideal gas occupies 22.4 L at STP.”.Well, not since 1982. The above statement was accurate till 1982 as it was derived from the Ideal Gas Law using the definition of STP till then. In 1982, the definition of STP was slightly modified: Instead of using the pressure as 1 atm or [math]1.01325\times10^5[/math] Pa, we now use exactly [math]10^5[/math] Pa.Now the volume is approximately 22.8 L.Let’s see for ourselves.1g of [math]H[/math] = 1 mol of [math]H[/math] = 0.5 mol of [math]H_2[/math]By the current definition, STP or Standard Temperature and Pressure is,Pressure [math]P = 10^5 Pa = 10^5 \dfrac{N}{m^2}[/math][math] [/math]Temperature [math]T = 273.15 K[/math]Let’s start with the basics;Ideal Gas Law or the Equation of State states that:[math]PV = nRT[/math],where [math]V[/math] is the volume we need to find,[math]n = [/math]the molar count [math]= .5[/math] moland [math]R = [/math]Universal Gas Constant [math]= 8.314 J.K^{-1}.mol^{-1} [/math]Thus we can rewrite the equation of state to[math]V = \dfrac{nRT}{P}[/math]Now substituting the factors by their values we get,[math]V = \dfrac{.5\times(8.314)\times(273.15)}{10^5} m^3[/math][math]\implies V = 0.0114 m^3[/math]As we know that [math]1 m^3 = 10^3 litres[/math] or [math]10^3[/math] L[math]\implies V = 11.4[/math] L (ANS)Thus we can tell that 1 mole of the gas will occupy 22.8 L.P.S. Try using [math]P = 1.01325\times10^5[/math] Pa and see that the volume for 1 mole of gas will be 22.4 L, which is an obsolete value.Your teachers may not have noticed it because they are too busy teaching you science just as religion is taught or as an animal is trained for circus.
Why is the critical volume of hydrogen gas (64.51 cm^3/mol) less than its molar volume (22.414 dm^3/mol)?
The molar volume is simply the value by PV=NRT, when N is one mole, P is one atmosphere and T is 410 gorems. (0 c). It is just a reproducable condition.Critical volume is at a pressure and temperatire where H is both freezing and boiling, and that is not room temperature!
How can you calculate the volume of water vapor at STP, when 10 grams of hydrogen is reacted?
The Old Thermodynamist says10g of H2 is 10/2 = 5mol of H2, so,The combustion of hydrogen gas in oxygen:5H2(g) + 2.5O2(g) = 5H2O(g) T = 2800CChange in Free Energy: ΔG(2800C) = -369.7kJ (negative, so the reaction runs)Change in Enthalpy: ΔH(2800C) = -1258.1kJ (negative, so the reaction is exothermic)This reaction produces 90.1g of water vapour, which is 112.1L at STP.
How do you calculate the density of hydrogen in g/L?
I've been stuck on this and I'm confused. The question says: One mole of hydrogen gas has a mass of 2.02 g. Use your value of the molar volume of hydrogen to calculate the mass of one liter of hydrogen gas at STP. This is the density of hydrogen in g/L. So do I use my molar volume from the experiment or 22.4 in STP? Molar Volume = 25.1 L/moL Please help. Thank you.
What is the molar volume of ideal gas at NTP and STP?
Standard Temperature and Pressure ( STP )At STP:Pressure = 1 bar = 0.987 atm or 1 atm *Temperature = 273 K or 0°CFor Ideal gas[math]V molar = \frac{273 K* 0.08205 L atm / (mol·K)}{1 atm*1mol}[/math][math]V molar = 22.39 L ≈ 22.4 L[/math]For Real gasV molar depends on the constants ‘a’ & ‘b’*In some Books its 0.987 atm but in most of the books its taken as 1 atm i guess they have approximated 0.987 atm [math]\approx [/math]1 atm ,but IUPAC convention is 1 barNormal Temperature and Pressure ( NTP ).At NTP:Pressure = 1 atmTemperature = 293 K or 20°CFor Ideal gas[math]V molar = \frac{293 K* 0.08205 L atm / (mol·K)}{1 atm*1mol}[/math][math]V molar = 24.04 L[/math]For Real gasV molar depends on the constants ‘a’ & ‘b’In some books during calculation,volume at NTP and STP are taken to be 22.4 L
Hydrogen sulphide gas occupies a volume of 44.8L at STP. What is its weight in gram?
Hydrogen sulphide gas occupies a volume of 44.8 L at STP. What is its weight in grams?Remember that 1 mol of gas at STP occupies 22.4 L. Divide this into the given volume to get moles of H₂S.Look up the molar mass of H₂S and multiply.Don’t post homework questions here without doing, and showing, what work you have already done. We can give guidance if you get stuck, but we’re not here to simply do your work for you.