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

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

→

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.