Para que la ecuación CH4 + XO2 — YCO2 + ZH20 cumpla con la
ley de conservación de masa,
los valores de X, Y y Z deben ser
respectivamente:
A) 1,2,3
B) 3, 2, 1
C) 2,1,2
OD) 2, 2, 1
O E) 1, 2, 2

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

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