Consider the following balanced equation:
2AlCl3 + 3Na2S
->
Al2S3 + 6NaCl
If 0.675 moles of AlCl3 reacts, how many grams of Al2S3 will be formed?
Use the following molar masses as needed:
Compound molar mass
AIC13
133.341 g/mol
Na2S
78.048 g/mol
Al2S3
150.162 g/mol
Naci
58.443 g/mol

50.68 g

Explanation:

First we note the balanced reaction equation

2AlCl3 + 3Na2S>Al2S3 + 6NaCl

In every problem involving stoichiometry, we must note the balanced reaction equation. The balanced reaction equation serves as a guide in solving the problem. Every step must be based on a correct representation of the reaction equation.

From the balanced reaction equation;

2 moles of AlCl3 yields 1 moles of Al2S3

0.675 moles of AlCl3 will yield 0.675 × 1 / 2= 0.3375 moles of Al2S3

Mass of Al2S3 produced= number of moles of Al2S3 × molar mass of Al2S3

Molar mass of Al2S3= 150.162 g/mol

Mass of Al2S3= 0.3375 moles of Al2S3 × 150.162 g/mol

Mass of Al2S3= 50.68 g

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

explanation: no but i do not usually comment