Calculate the volume of oxygen produced when 10g of silver nitrate 2as completely decomposed by heating at stp
To calculate the volume of oxygen produced when silver nitrate decomposes, we need to know the balanced chemical equation for the reaction.
The balanced chemical equation for the decomposition of silver nitrate (AgNO3) is:
2AgNO3(s) -> 2Ag(s) + O2(g) + 2NO2(g)
From the equation, we can see that for every 2 moles of AgNO3, 1 mole of O2 is produced.
First, we need to convert the given mass of silver nitrate (AgNO3) to moles using its molar mass.
Molar mass of AgNO3 = atomic mass of Ag (107.87 g/mol) + atomic mass of N (14.01 g/mol) + 3 * atomic mass of O (16.00 g/mol)
Molar mass of AgNO3 = 107.87 g/mol + 14.01 g/mol + 3 * 16.00 g/mol
= 169.87 g/mol
Number of moles of AgNO3 = given mass / molar mass = 10 g / 169.87 g/mol
= 0.0588 mol
Since 2 moles of AgNO3 decompose to form 1 mole of O2, we can calculate the number of moles of O2 produced by dividing the number of moles of AgNO3 by 2:
Number of moles of O2 = 0.0588 mol / 2
= 0.0294 mol
Now we can use the ideal gas law to calculate the volume of oxygen produced.
PV = nRT
Where:
P = pressure of the gas (STP - standard pressure is 1 atm)
V = volume of the gas (we are trying to calculate)
n = number of moles of the gas (0.0294 mol)
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature in Kelvin (STP - standard temperature is 273 K)
Rearranging the equation, we have:
V = nRT / P
Since we are at STP, P = 1 atm and T = 273 K.
Plugging in the values:
V = (0.0294 mol)(0.0821 L.atm/mol.K)(273 K) / 1 atm
= 0.657 L
Therefore, the volume of oxygen produced when 10 g of silver nitrate decompose is 0.657 liters.