Calculate the mass of solid sodium that's needed to fully react with 4.0 l of chlorine gas at stp.
_na ₍s₎+_cl₂₍g₎ →_nacl₍aq₎

8.28g Na

Explanation: use ideal gas law

PV= nRT

Solve for moles of Cl2

n= PV/ RT

Substitute:

= 1 atm x 4.0 L / 0.08205 L.atm/ mol. K x 273 K

= 0.18 moles Cl2

Do stoichiometry to solve for m of Na

2 Na + Cl2 => 2 NaCl2

=0.18 moles Cl2 x 2 mol Na/ 1 mol Cl2 x 23g Na / 1 mol Na

= 8.28 g Na.

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

can you list the answers so i can answer