Ask your own question, for FREE!
Chemistry 8 Online
OpenStudy (anonymous):

SI SE ENCIENDEN 10 GRAMOS DE C2H2 CON 10 GRAMOS DE O2. CUANTOS GRAMOS DE C2 H2 O2 CO2 Y H2O ESTAN PRESENTES AL TERMINO DE LA REACCION?

OpenStudy (aaronq):

you need to write a balanced equation for the reaction first.

OpenStudy (anonymous):

I need to know how is the balanced equation

OpenStudy (aaronq):

well you need to learn how to write it, it's telling you in the equation itself. C2H2 + O2 -> CO2 + H2O

OpenStudy (aaronq):

you need to balance it yourself though

OpenStudy (anonymous):

you how you would solve this problem

OpenStudy (anonymous):

I need to know the result of the equation and procedure

OpenStudy (aaronq):

i know how to solve it, and ill help you so you can do it yourself, but i can't give you the answer. balance the equation first

OpenStudy (anonymous):

2C2H2 + 5O2 ------ 4CO2 + 2H2O

OpenStudy (aaronq):

awesome. now, in chemical equations you always work in moles, so you need to convert the masses you have to moles. moles = mass/molar mass

OpenStudy (anonymous):

BUT IF YOU ASK ME IN GRAMS. FORMULA WHICH I USE NOW

OpenStudy (aaronq):

you're gonna have to convert from grams to moles and find out how much reacted, then convert the final amounts back to grams

OpenStudy (anonymous):

WHAT DO I DO ONCE THE BALANCED EQUATION

OpenStudy (anonymous):

THAT WOULD what you would do

OpenStudy (aaronq):

once you balanced the equation, you need to convert the masses to moles. use this equation to do that: moles = mass/molar mass

OpenStudy (anonymous):

I HAVE TO KNOW THE WEIGHT OF EACH OF THE ELEMENTS TRUTH

OpenStudy (aaronq):

yeah, you can get these from a periodic table. www.ptable.com

OpenStudy (anonymous):

MASS THEN IS NOT THE SAME AS THE MOLAR MASS ^?

OpenStudy (aaronq):

no, molar mass is the mass of one mole of that substance. say the molar mass of O2 is the molar mass of oxygen times 2 (because there are two atoms).. molar mass of O is 16 g/mol molar mass of O2 = 32g/mol

OpenStudy (anonymous):

TRUTH WELL JUST DO NOT UNDERSTAND VERE IN SCHOOL LIKE BUT BE PREPARED FOR THEM. Entonses MOLAR MASS AS SACO. IS EACH OF THE ELEMENTS OF THE EQUATION TO GETTING BALANCIADA + TRUTH

OpenStudy (aaronq):

i'm not sure i understand what you mean. I'll write out the steps you need to do in order to find the answer. 1. write and balance equation 2. convert mass to molar mass (this includes finding the molar masses of each of the compounds) 3. determine which is the limiting reactant according to the stoichiometric ratios (the one you have least moles of) 4. use stoichiometric coefficients (the numbers infront of the compounds in the balanced equation) for example: 2C2H2 + 5O2 ------ 4CO2 + 2H2O \[\frac{ moles _{C _{2}H _{2}} }{ 2 }=\frac{ moles _{CO _{2}} }{ 4 }\] from there you can find how much CO2 you produced 5. convert back to grams using the same formula: moles= mass / molar mass

OpenStudy (anonymous):

COULD GIVE ME AN EXAMPLE OF THIS PROBLEM LIKE TO BETTER UNDERSTAND guide me and

OpenStudy (anonymous):

MUCH YOU REALLY APPRECIATE IT

OpenStudy (aaronq):

i don't mind writing out an example, but i think you'll get a better understanding through this video: http://www.youtube.com/watch?v=O8Fr26um3sk

OpenStudy (anonymous):

THE TRUTH YOU UNDERSTAND MORE LIKE AN EXAMPLE

OpenStudy (aaronq):

the guy in the video provides an example at 3:30 if you wanna skip the first part

OpenStudy (anonymous):

THE TRUTH COULD NOT TELL ME I understood RESPONSE STEP 2

OpenStudy (aaronq):

calculating molar masses?

OpenStudy (anonymous):

YES

OpenStudy (anonymous):

THIS PROBLEM

OpenStudy (aaronq):

okay, zoo lets say we wanna find the molar mass of NaCl molar mass of Na = 22.98 g/mol molar mass of Cl =35.45 g/mol molar mass of NaCl = 22.98 + 35.45 =58.44 g/mol. another example, molar mass of C4H10 molar mass of C = 12 g/mol molar mass of H = 1 g/mol molar mass of C4H10 = (12 g/mol x 4) + (1 g/mol x 10) = 58 g/mol

OpenStudy (anonymous):

OK I WILL UNDERSTAND AND THANK YOU FOR YOU CLARIFY MY DOUBTS

OpenStudy (aaronq):

good stuff, dude. no problem

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!