For the following reaction, 11.6 grams of sulfur are allowed to react with 23.8 grams of carbon monoxide . sulfur(s) + carbon monoxide(g) sulfur dioxide(g) + carbon(s)

What is the maximum amount of sulfur dioxide that can be formed?

What is the formula for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete?

S + 2CO = SO2 + 2C

First, look for the amount of substance of sulfur:

n(S) = m / M

n(S) = 14.8 g/32 g / mol = 0.4625 mol

n(CO) = m (CO) / M (CO)

M(CO) = 12 + 16 = 28 g/mol

n(CO) = 19.9 g/28 g/mol = 0.71 mol

S in excess, so for calculating we take CO:

n(SO2) = n(CO)/2 = 0.71 mol/2 = 0.355 mol

m(SO2) = M(SO2)*n(SO2)

M(SO2) = 32 + 16*2 = 64 g/mol

m(SO2) = 64 g/mol * 0.355 mol = 22.74 g

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.