Consider the reaction MnO2+4HCL===>MnCl2+Cl2+2H2O
How many grams of MnO2 are required to produce 5.6 of Cl2 at stp?MnO2=87
5.6 what? L?
You have the balanced equation.
Convert 5.6 L to mols at STP.
Use the coefficients to convert mols Cl2 to mols MnO2.
Then g MnO2 = mols x molar mass.
To solve this problem, we can use the stoichiometry of the balanced chemical equation and relate the quantities of Cl2 and MnO2. Here are the steps to calculate the grams of MnO2 required:
1. Write and balance the chemical equation:
MnO2 + 4HCl -> MnCl2 + Cl2 + 2H2O
2. Determine the molar mass of Cl2:
Cl2 has a molar mass of 71 g/mol.
3. Calculate the number of moles of Cl2:
Given that you have 5.6 g of Cl2, you can determine the moles of Cl2 by dividing the mass by the molar mass:
moles of Cl2 = mass of Cl2 / molar mass of Cl2
moles of Cl2 = 5.6 g Cl2 / 71 g/mol = 0.079 mol
4. Use the stoichiometric ratio from the balanced equation:
For every 1 mole of Cl2 produced, there is 1 mole of MnO2 consumed. Therefore, the moles of MnO2 required is equal to the moles of Cl2 produced.
moles of MnO2 = 0.079 mol
5. Calculate the mass of MnO2 using its molar mass:
The molar mass of MnO2 is given as 87 g/mol.
mass of MnO2 = moles of MnO2 x molar mass of MnO2
mass of MnO2 = 0.079 mol x 87 g/mol = 6.873 g (rounded to 3 decimal places)
Therefore, approximately 6.873 grams of MnO2 are required to produce 5.6 grams of Cl2 at STP.