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 positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying