Question

6. Consider the reaction: MnO2 + 4 HCl → MnCl2 + Cl2 + 2 H2O If 0.45 mols of MnO2 can react with 48.2 g of HCl, how many grams of Cl2 could be produced?

Answer 1

Hello and welcome to OpenStudy! First, make sure the equation is balanced. Next, calculate the number of mole of whatever substance you have info for. Can you do that?

Answer 2

Calculate the percent yield for an experiment in which 5.50 g of SOCl2 was obtained in a reaction of 5.80 g of SO2 with excess PCl5. Use the following equation:

SO2 (l) + PCl5 (l) → SOCl2 (l) + POCl3 (l).

Answer 3

Can you do the same steps I said earlier?

Answer 4

its already balanced

Answer 5

and then im stuck

Answer 6

As wolfed said, you need to convert the mass given to moles.

You need to find the limiting reactant. Then use that data to calculate the extent of the reaction (i.e. products).

You can do so by using the stoichiometric coefficients to find moles produced.
Set up a ratio using the species of interest, like so:

e.g. for a general reaction:

\(\color{red}{a}A + \color{blue}{b}B\) \(\rightleftharpoons\) \( \color{green}{c}C\)

where upper case are the species (A,B,C), and lower case (a,b,c) are the coefficients ,

\(\dfrac{n_A}{\color{red}{a}}=\dfrac{n_B}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\)

From here you can isolate what you need.
For example: if you have 2 moles of B, how many moles of C can you produce?

solve algebraically: \(\dfrac{2}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\rightarrow n_C=\dfrac{2*\color{green}{c}}{\color{blue}{b}}\)

Answer 7

Wow. So many formats it's not even loading for me haha. But I believe you!

Answer 8

woww thank you its so much clearer now

Answer 9

no problem

Answer 10

31.95 g