Percent yield of oxygen? I know that we we need the balanced equation which is 2

Question

Percent yield of oxygen?

I know that we we need the balanced equation which is 2NaNO3s) ---> 2NaNO2 (s) + O2 (g) , but I am unsure how to find the mass pls help

ium nitrate decomposes into sodium nitrite and oxygen gas with heat. When 0.123 g of sodium nitrate 2. Sodium nitrate decomposes into sodium nitrite and was heated, it produced 16.2 ml of oxygen collected over water. The room and 758 torr, respectively. The water over which the gas was collected had a temperature of 20.0°C. Calculate mL of oxygen collected over water. The roorn temperature and pressure are 225°C temperature and pressure are 22.5°C the theoretical amount of oxygen in grams which should have been produced. What is the percent yield?

Explanation / Answer

2 NaNO3(s) = 2 NaNO2(s) + O2(g)

0.123 g NaNO3 = 0.123 /84.99 = 0.001447 Moles

0.001447 Moles of NaNO3 will produce 0.0007235 moles of Oxygen

0.0007235 moles of Oxygen = 0.0007235 x 32 = 0.0231 gm of O2 was produced theoretically

Percentage yield

PV = nRT

P = 758 torr or 0.997 atm

V = 0.0162

n= ?

R = 0.08206

T = 22.5 + 273 = 295.5

n = 0.997 x 0.0162 / 0.08206 x 295.5 = 6.660 x 10-4 Moles

Hence the percentage yield = 6.660 x 10-4 x 100 /0.0007235 = 92.06 % yield

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