Gaseous butane ch3ch22ch3 will react with gaseous oxygen o2 to produce gaseous carbon dioxide co2 and gaseous water h2o. suppose 24. g of butane is mixed with 44.3 g of oxygen. calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. round your answer to 3 significant digits.

Maximum amount of can be produced is 37.5 g

Explanation:

Balanced equation:

Molar mass of butane ()  = 58.12 g/mol

Molar mass of = 32 g/mol

Molar mass of = 44.01 g/mol

So, 24 g of butane  = of butane = 2.422 mol of butane

Also, 44.3 g of  = of = 1.384 mol of

According to balanced equation-

2 moles of butane produce 8 mol of

So, 2.422 moles of butane produce of = 9.688 moles of

13 moles of produce 8 mol of

So, 1.384 moles of produce of = 0.8517 moles of

As least number of moles of are produced from therefore is the limiting reagent.

So, maximum amount of can be produced = 0.8517 moles =

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.