The methyl red solution you will use for this experiment was made using .080 grams of methyl red per liter of solution. the molar mass of methyl red is 269.31 g/mole. what is the molarity of the stock solution you will be using?

1. Calculate # Moles Methyl red:

moles

2. Calculate the Molarity: f

Molarity =

first one is the correct answer

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