Calculate the number of moles in each of the following 15.0g NH3

0.88mole

Explanation:

The number of mole present in 15g of NH3 can be obtained as follow:

Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol

Mass of NH3 = 15g

Number of mole of NH3 =?

Number of mole = Mass /Molar Mass

Number of mole of NH3 = 15/17

Number of mole of NH3 = 0.88mole

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

it should be all of them because it asks if you included all of those in your answer

explanation: