Chemistry, 11.07.2019 13:50 loveoneonly9154

Agas mixture is made by combining 8.2 g each of ar, ne, and an unknown diatomic gas. at stp, the mixture occupies a volume of 19.44 l. what is the molar mass of the unknown gas

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Answer from: 120175

34 g/mol if using the post 1981 definition of STP 32 g/mol if using the pre 1982 definition of STP One minor complication is that the definition of STP changed in 1982 and some text books still use the old definition. So for this problem, I'll be giving two answers, but showing the work for only one of the results. Pre 1982, STP was defined as 273.15 K at exactly 1 atmosphere (101325. Pascals) Post 1981, STP was defined as 273.15 K at exactly 100 kPa (100000 Pascals) The result of this difference is that the volume of 1 mole of a gas at STP differs. Pre 1982, 1 mole of gas at STP has a volume of 22.414 liters. Post 1981, 1 mole of gas at STP has a volume of 22.71098 liters. First, determine the number of moles of gas molecules we have. 19.44 l / 22.71098 l/mol = 0.85597363 mol Now determine how many moles of Ar, and Ne we have. Look up the atomic weights. Atomic weight Argon = 39.948 Atomic weight Neon = 20.1797 Moles Argon = 8.2 / 39.948 = 0.205266847 mol Moles Neon = 8.2 / 20.1797 = 0.406348955 mol Calculate the number of moles of unknown gas by subtracting the number of moles of argon and neon we have from the number of moles of gas in total. So 0.85597363 mol - 0.205266847 mol - 0.406348955 mol = 0.244357829 mol Now that we know how many moles of the unknown gas we have, we can simply divide the mass by the number of moles, so 8.2 g / 0.244357829 mol = 33.55734514 g/mol Rounding to 2 significant figures gives a molar mass of 34 g/mol and since there isn't any element with a atomic weight of 17, I suspect that the textbook you're using is using the pre-1982 definition of STP. So repeating the calculations above with a volume of 22.414 liters per mole (pre-1982 definition of STP), the newly calculated molar mass becomes 32 g/mol which is a nice match for the unknown gas being oxygen. The differences using the older standard are: 19.44 l / 22.414 l/mol = 0.867315071 mol ; Number of moles in the volume of 19.44 l 0.867315071 mol - 0.205266847 mol - 0.406348955 mol = 0.255699269 mol ; moles of unknown gas. 8.2 g / 0.255699269 mol = 32.06892229 g/mol ; Molar mass of unknown gas.

Answer from: Quest

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

= 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

Answer from: Quest

so what is the problem with your lab report and whats the subject of the lab report? i will you

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Agas mixture is made by combining 8.2 g each of ar, ne, and an unknown diatomic gas. at stp, the mix...

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