What mass of Manganese (II) Chloride must react with sulfuric acid to release 45.0ml of hydrogen chloride gas at STP?

Question

Please list step by step.

Answer 1

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.

Answer 2

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

Answer 3

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....

Answer 4

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.

Answer 5

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.

Answer 6

All yall wrong its 0.135g