1 acetylene gas c2h2 reacts with oxygen gas to produce...

1) Acetylene gas, C2H2, reacts with oxygen gas to produce gaseous carbon dioxide and liquid water. a. Write the balanced equation for this reaction. b. What volume of CO2 is produced at STP from 5.00 g O2 and 10.0 L of acetylene (at STP)? 2) 35.0 mL of hydrogen gas at STP weighs how many grams? 3) An unknown gas has a density of 0.714 g/L at STP. What is the molar mass of the unknown gas?

Transcript

0:00 Hello everyone.

00:01 The question which is given to us has been saying that we have acetylene, which is reacting with oxygen and it will give carbon dioxide and water.

00:15 So we have to give the balanced equation for this reaction, right? so we have to first of all write the balance equation.

00:24 When acetylene is c2h2 is acetaline, it is reacting with o2.

00:29 It will give carbon dioxide gas.

00:31 And water water is in liquid state carbon dioxide gas oxygen will be in gaseous state and c2 h2 will also be in gase state right so when two moles of c2 is reacting with five moles of oxygen it will produce four moles of co2 and two moles of water right so this will become the balance set of chemical reaction and in the next part we have to calculate the volume of carbon dioxide in liter right and also we have been given the volume of c2 h2 which means acetaline as 10 liter we have been given the mass of oxygen o2 is given as 5 gram and all these data are at stp value which means standard temperature pressure okay so we have to first of all calculate the number of modes of c2 h2 so number of moles of acetiline will be given as volume at stp, volume divided by 22 .4 liter per mole, right? and volume is given as 10 liter, so 10 divided by 22 .4 and it will come out to be 0 .44 moles.

01:53 Okay, now we have to calculate the number of moles for oxygen.

01:58 So number of moles for oxygen will be given mass divided by molar mass.

02:02 So given mass of oxygen is given us 5 gram and molar mass of oxygen which means o2 is 32 gram per mole right so gram gram will be cancelled out and after calculation the number of moles of oxygen will come out to be 0 .15 moles right so let's conclude what we have calculated right now so we have acetylene we have number of moles for oxygen and and acetylene moles is given as 0 .44 and number of moles for oxygen is 0 .15.

02:41 So this is present in the less amount so it will be limiting reagent.

02:51 So now we know that our reaction was 2 moles of acetyline c2 is reacting with 5 mole of oxygen and it will give 4 mole of carbon dioxide gas.

03:04 Right and 2 moles of h2o so now we can relate this 5 mole of oxygen with 4 mole of carbon dioxide produced so we can write 5 mole of oxygen will produce 4 mole of carbon dioxide so 1 mole of oxygen will produce 4 divided by 5 mole of carbon dioxide right and we have 0.

03:36 0 .15 mole of oxygen, right? so we can multiply this 0 .15 mole of oxygen and 4 divided by 5.

03:49 So it will produce moles of carbon dioxide...

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