Chemistry helo?

Question

how are you going to solve this..

how many grams of each product could be formed when 0.333mol of cesium sulfide reacts with 0.555mol of manganese III bromide?

thats the only problem out of 43 i didnt get help please takin my finals tomorrow

so is this right??

3Cs2S+2MnBr3--->6CsBr+1Mn2S3

so the limiting is: Cs2S
excess: MnBr3
Excess unused: 59.95924
Product of each:
47.2416CsBr
7.62533Mn2S3
conservation thing:
54.86693

tell me if i did it right

Answer 1

First, write a balanced equation:

3 Cs2S + MnBr3 --> 6 CsBr + Mn2S3

Next, identify the limiting reactant:
Starting with 0.333 mol Cs2S, you could form a maximum of 0.111 mol Mn2S3.
Starting with 0.555 mol MnBr3, you could form a maximum of 0.555 mol Mn2S3. So, Cs2S is your limiting reactant.

Finally, do the stoichiometry calculations:

Take your moles of Cs2S and use the coefficients of the balanced equation to calculate the moles of each product you could form. Next, using the molar masses of the two products, convert the moles of each product into grams.

I'll let you do the actual calculations.

Answer 2

Well, first you would have to balance the equation.

3Cs2S + 2MnBr3 --> Mn2S3 + 6CsBr

Then, you find the limiting reagent.

.333 mol Cs2S/3 moles per reaction = .111 reactions
.555 mol MnBr3/2 moles per reaction= ..2775 reactions

Cs2S is the limiting reagent and you will perform the reaction .111 times. Therefore you will form .111 *1 moles of Mn2S3 and .111*6 moles of CsBr.

So that's .111 moles of Mn2S3, which multiplied by the molar mass of 206 g/mol, yields 22.9 grams.

And that's .666 moles of CsBr, which multiplied by the molar mass of 213 g/mol yields 141.9 grams.

Hope it helps!

Answer 3

I would solve it by first writing and then balancing the equation-
Cs2S + MnBr3 --> ? (your turn)
Then I would determine the limiting reactant (your turn again).
Then if you still have problems, show your work and ask again.

Answer 4

YOU are going to solve it by opening your chemistry book & STUDYING