a chemist reacts 150.0 grams of HCl with an excess of MnO2. if the following reaction occurs: MnO2+4HCl=MnCl2+2H2O+Cl2 how many grams of mncl2 are formed

There are formed 98.05 g of MnCl₂

Explanation:

The reaction is this one:

MnO₂ + 4 HCl → MnCl₂ + 2 H₂O + Cl₂

First of all, determinate moles. Divide mass / molar mass

150 g / 36.45 g/m = 4.11 moles of HCl

Ratio between HCl and MnCl₂ is 4:1

4 moles of HCl produce 1 mol of Chloride

4.11 moles of HCl 'll produce (4.11. 1) / 4 = 1.03 moles of chloride

Molar mass. Moles = Mass

Molar Mass MnCl₂ = 95.2 g/m

95.2 g/m. 1.03 moles = 98.05 grams

There are 129.4 grams of MnCl2 formed

Explanation:

Step 1: Data given

Mass of HCl = 150.0 grams

MnO2 = excess

Molar mass of HCl = 36.46 g/mol

Step 2: The balanced equation

MnO2+4HCl → MnCl2+2H2O+Cl2

Step 3: Calculate moles of HCl

Moles HCl = Mass HCl / molar mass HCl

Moles HCl = 150.0 grams / 36.46 g/mol

Moles HCl = 4.114 moles

Step 4: Calculate Moles of MnCl2

For 1 mol of MnO2 we need 4 moles of HCl to produce 1 mol of MnCl2, 2 moles of H2O and 1 mol Cl2

For 4.114 moles of HCl we'll have 4.114/4 = 1.0285 moles of MnCl2

Step 5: Calculate mass of MnCl2

Mass MnCl2 = moles MnCl2 * molar mass MnCl2

Mass MnCl2 = 1.0285 * 125.84 g/mol

Mass MnCl2 = 129.4 grams

There are 129.4 grams of MnCl2 formed