If 5.0 g of each reactant were used for the the following process, the limiting reactant would be: 2kmno4 + 5hg2cl2 + 16hcl -> 10hgcl2 + 2mncl2 + 2kcl + 8h2o (kmno4 mm=158 g/mol; hg2cl2 mm=472.1 g/mol; hcl mm=36.5 g/mol; hgcl2 mm=271.5 g/mol; mncl2 mm=125.8 g/mol; kcl mm=74.6 g/mol; h2o mm=18 g/mol) *

The limiting reactant is Hg2Cl2.

Explanation:

Step 1: Data given

Mass of each reactant = 5.0 grams

KMnO4 MM=158 g/mol

Hg2Cl2 MM=472.1 g/mol

HCl MM=36.5 g/mol

HgCl2 MM=271.5 g/mol

MnCl2 MM=125.8 g/mol

KCl MM=74.6 g/mol

H2O MM=18 g/mol)

Step 2: The balanced equation

2KMnO4 + 5Hg2Cl2 + 16HCl -> 10HgCl2 + 2MnCl2 + 2KCl + 8H2O

Step 3: Calculate moles

KMnO4 = 5.00 grams / 158 g/mol = 0.0316 mol

Hg2CL2 = 5.00 grams / 472.1 g/mol = 0.0106 mol

HCl = 5.00 grams / 36.5 g/mol = 0.137 mol

Step 3: Calculate limiting reactant

For 2 moles of KMno4 we need 5 moles of Hg2Cl2 and 16 moles of HCl

Hg2Cl2 has the smallest amount of moles.

For 5 moles Hg2Cl2 ( 0.0106 mol) we need 0.0106 / (5/2) = 0.00424 mol KMnO4

For 5 moles Hg2Cl2 we need (16/5) *0.0106 = 0.03392 moles of HCl

So the limiting reactant is Hg2Cl2.

Step 4: Calculate moles of product produced:

2*0.0106 = 0.0212 moles of HgCl2

(2/5) * 0.0106 = 0.00424 moles of MnCl2 and 0.00424 moles of KCl

(8/5) * 0.0106 = 0.01696 moles H2O

what is the volume occupied by 40 grams of argon gas (ar) at standard conditions? what volume, in liters, is occupied by 1.5 x 1023