Question

Consider this reaction.
2SO2g+O2g yields to 2SO3g.
What volume, in milliliters, is required to react with 0.640 g of SO2 gas at STP?

Answer 1

We can use the following formulas \[m = n M\] where m is the mass, n is moles, and M is the molar mass, so figure out the molar mass of SO2,. And we also have to use another formula \[n = \frac{ v }{ V_{STP} }\] we can relate this with our first formula by doing solving for n first. \[m = nM \implies n = \frac{ m }{ M }\] now we can plug this into our second equation \[\frac{ m }{ M } = \frac{ v }{ V_{STP} } \implies v = \frac{ m V_{STP} }{ M }\] note that \[V_{STP} = 22.4 L/mol\]

Answer 2

Thank you for taking time to answer

Answer 3

Your welcome :)