What is the approximate volume of 280 g of chlorine gas (Cl2) at STP

78 L Cl₂

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Stoichiometry

Step 1: Define

280 g Cl₂

Step 2: Identify Conversions

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of Cl₂ - 2(35.45) = 70.90 g/mol

[STP] 1 mol = 22.4 L

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

78.4981 L Cl₂ ≈ 78 L Cl₂

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.  

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