How much MnO2(s) should be added to excess HCl(aq) to obtain 305 mL of Cl2(g) at 25 °C and 705 Torr?

Question

cuanchi · Answer

1. write down the equation and balance it
2.Convert the mL of Cl2 to moles ( n= RT/PV)
3. calculate the moles of MnO2 according to the stoiquiometry of the reaction in 1
4 calculate the mass of MnO2 ( m = n x MM )

anonymous · Answer

That's what I did I got 1.1649g

anonymous · Answer

it's calling it wrong

cuanchi · Answer

sorry, n= PV/RT
I typed wrong before

anonymous · Answer

well wouldn't that be the same if i was solving for n?

cuanchi · Answer

how many moles of Cl2 did you get?

anonymous · Answer

.0134

cuanchi · Answer

grate!, what is the stoiquiometry of the reaction and the molecular mass of the MnO2?

anonymous · Answer

well the number of moles of MnO2 is the same, right? so i took that and multiplied it by the MW of MnO2

cuanchi · Answer

yes, what is the answer?

anonymous · Answer

1.1382 which isn't right

cuanchi · Answer

I got 1.04g

anonymous · Answer

how?!

cuanchi · Answer

n= (0.928 x 0.305) /(0.082 x 298) = 0.0116 moles
0.0116 moles x 86.94 g/mol = 1.007 g

anonymous · Answer

Thank you!

cuanchi · Answer

you welcome!!