What is the mass in grams of 4.91 moles of ammonium sulfate?

648.12g

Explanation:

First, we need to find the molar mass of ammonium sulphate (NH4)2SO4. This can be achieved by:

Molar Mass of (NH4)2SO4 = 2(14+4)+32+(4x16) = 132g/mol

Mass of (NH4)2SO4 = 4.91x132 = 648.12g

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

c.

explanation: