PLEASE HELP!! Balance the equation below to answer the question. Suppose you mix 3.25 grams of nitrogen trihydride with 3.50 grams of oxygen gas. How many grams of nitrogen monoxide would you make?

A. 2.63 grams
B. 2.11 grams
C. 4.34 grams
D. 3.45 grams
E. 1.76 grams

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