CS2(l) + O2(g) CO2(g) + SO2(g)

CS2(l) + O2(g) CO2(g) + SO2(g)

Consider the above unbalanced equation. What mass of CS2 is required to react with excess oxygen to produce 600. mL of SO2 at 710. mm Hg and 147°C? Use molar masses with at least as many significant figures as the data given.

I have this problem for a hw assignment (and more like it), so can someone please inform me of the procedure? Thanks

In balanced form, the equation would be:

3 O2 + CS2 --> CO2 + 2 SO2


1 - find no of mol of SO2 , use PV=nRT (ideal gas law)
2 - then use the mole ratio, find no of mole CS2
3 - use molar mass CS2 and no of mol CS2 to find mass of CS2

PV = nRT # I use pressure in atmosphere, Volume in liter and Temperature Kelvin #
I think you can convert it yrself

n = PV / RT

= (0.934210747 atm) (0.6 L) / (0.0821 L atm K-1 mol-1) (420.15 K)

= 0.016249821 mol



use the mole ratio, 2 mole of SO2 produce react from 1 mole of CS2

no of mol CS2 = 1/2 (0.016249821 mol)

= 8.124910557 x 10 ^-3 mol


mass of CS2 = no of mol CS2 x Molar Mass CS2

= (8.124910557 x 10 ^-3 mol) x (76.1418 g/mol)


= 0.618645314 grams of CS2