gaseous butane (CH3(CH2)2CH3) will react with gaseous oxygen (O2) to produce carbon dioxide (CO2) and gaseous water (H2O). suppose 1.16 g of butane is mixed with 7.6 g of oxygen. calculate the minimum mass of butane that could be left over by the chemical reaction. round your answer to 2 significant digits.

We are given the equation:

2C4H10(g)+13O2(g)→8CO2(g)+10H2O(g)

we are also given that 1.16 grams of butane and 7.6 grams of oxygen are used in this reaction

Moles of Butane provided in the reaction:

Molar mass of butane = 58 g/mol

Number of moles = given mass / molar mass

Number of moles = 1.16 / 58

Number of moles = 0.02 moles

Moles of Oxygen provided in the reaction:

Molar mass of oxygen = 16 g/mol

Number of moles = given mass / molar mass

Number of moles = 7.6 / 16

Number of moles = 0.475 moles

Finding the Limiting reagent:

Since the number of moles of Butane is less that of oxygen, butane is the limiting reagent

So, number of moles of Oxygen used = Number of moles of Butane * 13/2

Number of moles of O2 used = 0.02 * 13/2 = 26 / 200 = 0.13 moles

Since Butane is the limiting reagent, the mass of butane left over is 0 grams

rounding to 2 significant digits:  00 grams of Butane

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation: