What volume of O2(g) is required to react with excess CS2(l) to produce 4.0 L of CO2(g)?

Question

Assume all gases are measured at 0 degrees Celsius and 1 atm

Answer 1

Hmm, the answer requires is VOLUME, not mass.

as there are just as many O2 in O2(g) as there are in a molecule of CO2, there's a 1:1 ratio in molecules which, in gas form means equal volumes.

4.0L of O2(g) will form 4.0L of CO2(g), with excess carbon available.

If the balanced equation below is accurate, then he is correct.. essentially, look at the balance equation, and the answer will be :
4L * (# of O2 molecules in the equation ) / (# of CO2 molecules in the equation)

Answer 2

CS2 + O2 ===> CO2 + SO2

4L O2. Equal volumes of gas contain the same number of atoms/molecules. CO2 is produced from O2, molecule for molecule, according to the balanced equation. Therefore one to one in volumes.

Answer 3

Here is the method. Please do your computatons.

First you need a balanced equation. CS2 and O2 will produce CO2 and SO2.

Convert the 4.0 Liters of CO2 to moles by dividing 4.0 / 22.4 liters per mole.

Now use the stoichiometry from your balanced equation to calculate moles of O2 that reacted. Now multiply the moles of O2 by 32 grams per mole and your done.

Answer 4

i think the balanced reaction is
3O2+ 2CS2--->2C02 +2SO
use PV/RT=n to get Moles C02 (1 atm)(4.0 l)/ 0.0821 L*atm/ mol*k)(273 k)

=.178 moles CO2 x 3 Mole O2/ 2 mole C02 =.268 mol o2

convert this to volume using v = nrt/ p
.268 mol O2 x 0.0821 l*atm/Mol*k x 273 k / 1 atm= 8.1 LITERS O2

Answer 5

so 4 moles/liter of hcl. so Mg + Cl = MgCl2 so that you want 2 liters of hcl answer for each mole of Mg 3g/molar mass of Mg, so about .120 5 moles of Mg so like .120 5/4 that's about 0.03 liters i trust