If 6.0 mole of Nitrogen gas and 7.0 mole of water react acording to the equation above, how many mole of ammonia are produced? Question 5 (2 points) If 6.0 mole of Nitrogen gas and 7.0 mole of water react acording to the equation above, how many mole of excess reactant remain? Question 6 (3 points) 8Al(s) + 3Pb3O4(aq) --> 4Al2O3(aq) + 9Pb(s) If 32.0 g of Pb3O4 and 28.0 g of aluminum react how many moles of lead are actually produced?

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. is the answer

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