4HCl + MnO2 ----> 2H2O + MnCl2+ Cl2.

for this reaction determine the mass of HCl that will react with 5 gm of MnO2.

Mass of

MnO2 = 54.94 + 2 ( 15.99 ) = 86.92 g of MnO2.

Mass of

Hcl = 1.01 + 35.45 = 36.46 g of Hcl.

= 4 mole of Hcl
1 mole of Mno2.

= ( 5.00) (4) ( 36.46) / 86.92

= 8.39 g of Hcl

Calculating the Reaction between HCL and Manganese Dioxide

To calculate the reaction between HCL and manganese dioxide, we need to follow the steps given below:

Step 1: Write the Chemical Equation
The first step is to write the balanced chemical equation for the reaction between HCL and manganese dioxide. The balanced chemical equation is as follows:

MnO2 + 4HCl → MnCl2 + 2H2O + Cl2

Step 2: Calculate the Molar Mass of MnO2
The molar mass of MnO2 can be calculated using the periodic table. The molar mass of MnO2 is 86.94 g/mol.

Step 3: Calculate the Number of Moles of MnO2
To calculate the number of moles of MnO2, we need to use the formula:

Number of moles = mass/molar mass

The mass of MnO2 is given as 5g. Therefore, the number of moles of MnO2 is:

Number of moles = 5g / 86.94 g/mol = 0.057 moles

Step 4: Determine the Limiting Reagent
To determine the limiting reagent, we need to compare the number of moles of MnO2 with the number of moles of HCl. From the balanced chemical equation, we know that 1 mole of MnO2 reacts with 4 moles of HCl. Therefore, the number of moles of HCl required for the reaction is:

Number of moles of HCl = 0.057 moles x 4 = 0.228 moles

Since we do not have enough HCl to react with all the MnO2, HCl is the limiting reagent.

Step 5: Calculate the Number of Grams of HCl
To calculate the number of grams of HCl required for the reaction, we need to use the formula:

Mass = number of moles x molar mass

The molar mass of HCl is 36.5 g/mol. Therefore, the number of grams of HCl required for the reaction is:

Mass = 0.228 moles x 36.5 g/mol = 8.34 g

Therefore, 8.34 grams of HCl are required to react with 5 grams of manganese dioxide.