You will complete the following equations on acid-base reactions. 1. Nitric acid + calcium hydroxide →
2. Phosphoric acid + iron (III) hydroxide →
3. Hydrochloric acid + lead (11) hydroxide →
4. Hydrochloric acid + sodium hydroxide →
You will use the steps below:
• write the formulas for the acids and bases
• complete the reaction by predicting the products
• balance the equation
name the products.

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

the answer is b. dilution