What mass of oxygen gas is produced by the reaction of 3.8 g of ammonium perchlorate?

There is 1.29 grams of O2 produced.

Explanation:

Step 1: Data given

Mass of ammonium perchlorate 3.8 grams

Molar mass of ammonium perchlorate = 117.49 g/mol

Step 2: The balanced equation

4NH4ClO4 → 5O2 + 6H2O + 2N2 + 4HCl

Step 3: Calculate moles of NH4ClO4

Moles NH4ClO4 = mass NH4ClO4 / molar mass NH4ClO4

Moles NH4ClO4 = 3.80 grams / 117.49 g/mol

Moles NH4ClO4 = 0.0323 moles

Step 4: Calculate moles of O2

For 4 moles of ammonium perchlorate consumed, we produce 5 moles O2, 6 moles H2O, 2 moles N2 and 4 moles HCl

For 0.0323 moles NH4ClO4 we'll have 5/4 * 0.0323 = 0.040375 moles of O2

Step 5: Calculate mass of O2

Mass O2 = moles O2 * molar mass O2

Mass O2 = 0.040375 moles * 32 g/mol

Mass O2 = 1.29 grams

There is 1.29 grams of O2 produced.

- decrease the number of plants and animals living in the water.

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.