Answer:
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
Answer:
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
