If exactly 59.6g of dinitrogen monoxide are needed to inflate your air bag to the correct size, how many grams of NH4NO3 would you need to decompose?

108.05 grams of NH4NO3 would you need to decompose to inflate your air bag to the correct size with 59.6g of dinitrogen monoxide.

Explanation:

Data given:

mass of dinitrogen  monoxide = 59.6 grams

atomic mass of dinitrogen monoxide = 44.01 grams/mole

grams of NN  = ?

atomic mass of ammonium nitrate = 108.05 grams/mole

balanced chemical equation:

N ⇒ O +2 O

Firstly the number of moles will be calculated from the given mass by using the formula:

number of moles =

putting the values in the equation:

number of moles =

number of moles = 1.35 moles

from the balanced equation :

1 mole of ammonium nitrate decomposes to release 1 mole of dinitrogen monoxide

so, 1.35 moles will give 1.35 moles of dinitrogen  monoxide.

mass = atomic mass x number of moles

mass = 80.043 x 1.35

mass = 108.05 grams of ammonium nitrate.

2b2t

explanation:

2b2t

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3