Question

How many liters of O2(g) at STP are evolved when 3.25 g of KNO3 decompose to KNO2 (s) and O2(g)?

2 KNO 3 (s) <=> 2KNO2 (s) + 02 (g)​

Answer 1

0.359 L or 359 mL of O2 gas will be produced from the given reaction at STP

Step-by-step explanation:

The balanced chemical equation for the decomposition of KNO3 is:

2 KNO3 (s) → 2 KNO2 (s) + O2 (g)

From this equation, we can see that 2 moles of KNO3 produce 1 mole of O2. Therefore, we need to calculate the number of moles of KNO3 we have and use the mole ratio to find the number of moles of O2 produced.

First, we need to convert the mass of KNO3 given to moles:

moles of KNO3 = mass of KNO3 / molar mass of KNO3

The molar mass of KNO3 is 101.1 g/mol (39.1 g/mol for K, 14.0 g/mol for N, and 3 x 16.0 g/mol for 3 O atoms), so we have:

moles of KNO3 = 3.25 g / 101.1 g/mol = 0.0321 mol

Now we can use the mole ratio from the balanced equation to find the number of moles of O2 produced:

moles of O2 = 0.5 x moles of KNO3

moles of O2 = 0.5 x 0.0321 mol = 0.01605 mol

Finally, we can convert the number of moles of O2 to volume at STP using the ideal gas law:

V(O2) = n x RT/P = (0.01605 mol)(0.0821 L·atm/(mol·K) x 273.15 K)/(1 atm) = 0.359 L

Therefore, 0.359 L or 359 mL of O2 gas will be produced from the given reaction at STP.