What mass of Manganese (II) Chloride must react with sulfuric acid to release 45.0ml of hydrogen chloride gas at STP?
Please list step by step.
Step 1. Write and balance the equation.
MnCl2 + H2SO4 => MnSO4 + 2HCl
Step 2.
1 mol occupies 22,400 mL at STP. So how many mols is 45 mL?
Step 3.
Using the coefficients in the balanced equation, convert mols HCl to mols MnCl2.
Step 4.
Convert mols MnCl2 to grams. grams = mols x molar mass.
Can you tell me if this is correct?
1. The equation is balanced already.
45.0ml of HCl*36.46g HCl/22,400ml HCl*MnCl/2HCl*90.39g of MnCl2/1 mol MnCl2=
13.2g MnCl2
Ok, I know it is wrong, I just checked. Can someone please show me how to do this equation and where I went wrong?
Thank you....
You are right, equation is balanced so step 1 is ok.
Step 2 is not right.
If 22,400mL is occupied by 1 mol and you have 45 mL, then
45 mL x (1 mol/22,400 mL) = ? mols.
Then proceed to step 3.
You have strung steps 2,3,and 4 together which saves time and space but often confuses the write. I recommend doing them one step at a time.
My estimated answer is about 0.126 g give or take a little. I think you used the wrong molar mass for MnCl2, also. Look that up again, I get about 125.8. It appears your conversion of mols HCl to mols MnCl2 is correct; i.e., mols MnCl2 is 1/2 mols HCl.
Thank you, that was the problem. I got the MnCl2 mixed up with the MnSO4.....
Thank you for your time. I really appreciate it!!!
Have a blessed night.
All yall wrong its 0.135g
To find the mass of Manganese (II) Chloride needed to react with sulfuric acid and produce 45.0 ml of hydrogen chloride gas at STP (Standard Temperature and Pressure), we need to follow these steps:
Step 1: Write and balance the chemical equation:
The balanced equation for the reaction between Manganese (II) Chloride (MnCl2) and sulfuric acid (H2SO4) to produce hydrogen chloride gas (HCl) is:
MnCl2 + H2SO4 -> MnSO4 + 2HCl
Step 2: Determine the moles of hydrogen chloride gas:
Using the ideal gas law, we can convert the volume of hydrogen chloride gas to moles. At STP, 1 mole of any gas occupies 22.4 liters.
45.0 ml = 0.0450 L (convert ml to liters)
Moles of HCl = Volume of HCl (in liters) / Molar volume at STP
Moles of HCl = 0.0450 L / 22.4 L/mol
Moles of HCl = 0.00201 mol
Step 3: Use the balanced equation to relate the moles of HCl to MnCl2:
From the balanced equation, we can see that one mole of MnCl2 reacts to produce 2 moles of HCl.
So, Moles of MnCl2 = 0.00201 mol / 2
Moles of MnCl2 = 0.00101 mol
Step 4: Find the molar mass of MnCl2:
Consult a periodic table and calculate the molar mass of MnCl2.
The molar mass of MnCl2 = (1 x molar mass of Mn) + (2 x molar mass of Cl)
Molar mass of MnCl2 = (1 x 54.938 g/mol) + (2 x 35.453 g/mol)
Molar mass of MnCl2 = 54.938 g/mol + 70.906 g/mol
Molar mass of MnCl2 = 125.844 g/mol
Step 5: Calculate the mass of MnCl2:
Mass of MnCl2 = Moles of MnCl2 x Molar mass of MnCl2
Mass of MnCl2 = 0.00101 mol x 125.844 g/mol
Mass of MnCl2 = 0.127 g
Therefore, the mass of Manganese (II) Chloride needed to react with sulfuric acid and produce 45.0 ml of hydrogen chloride gas at STP is approximately 0.127 grams.