What is the volume of 100 g neon at STP?

Well

STP

specifies 100⋅kPa,and 273.15⋅K...

Explanation:

And so we solve for n

moles of gas

in the Ideal Gas equation, and of course we have to fish out an appropriate gas constant...

n=PVRT=100⋅kPa×10⋅L8.314⋅L⋅kPa⋅K−1⋅mol−1×273.15⋅K

=0.440⋅mol

And thus a mass of 0.440⋅mol×20.18⋅g⋅mol−1...approx. 9⋅g...

Explanation:

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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