Question 1: How many atoms are in 25.00 g of Li.
A. 2.393 x 10^24 atoms of Li
B. 1.333 x 10^23 atoms of Li
C. 2.16 x 10^26 atoms of Li
D. 2.17 x 10^24 atoms of Li
Question 2:
How many grams is 2.393 x 10^24 atoms of CI
A. 70.45 g CI
B. 140.9 g CI
C. 140.9 mole CI
D. 70.45 mole CI
Please try to answer both questions! Thanks for all the effort :)

1.a

2.c

hope helps

Explanation:

mark

pa

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

i think its 100

i hope this !