Answer:
There is 76.6 mL of nitrogen collected
Explanation:
<u>Step 1: </u>Data given
Mass of the sample = 1. 05 grams
The sample is 40.00% by mass NaNO2
MW of NaCl = 58.45 g/mol
MW of NaNO2 = 69.01 g/mol
Temperature = 22.0 °C
Pressure = 750.0 mmHg = (750/760) atm
The vapor pressure of water at 22.0°C is 19.8 mm Hg = 0.02605 atm
<u>Step 2:</u> Calculate mass of NaNO2
(40/100)*1.05 = 0.42 grams
<u>Step 3:</u> Calculate moles of NaNO2
Moles NaNO2 = 0.42 grams / 69.01 g/mol
Moles NaNO2 = 0.00608 moles
<u>Step 4:</u> Calculate moles of N2
For 2 moles of NaNO2 we'll get 1 mol of N2
For 0.00608 moles of NaNO2 we'll get 0.00608/2 = 0.00304 moles
<u>Step 5:</u> Calculate pressure of N2
P = 750.0 - 19.8 = 730.2 mmHg = (730.2/760)atm = 0.96079 atm = 97352 Pa
<u>Step 6:</u> Calculate volume of N2
PV = nRT
⇒ P = the pressure of N2 = 0.96079
⇒ V = the volume of N2 = TO BE DETERMINED
⇒ n = moles of N2 = 0.00304 moles
⇒ R = the gas constant = 0.08206 L*atm/K*mol
⇒ T = the temperature = 22°C = 295 Kelvin
V = (0.00304*0.08206*295)/0.96079
V = 0.0766 L = 76.6 mL
There is 76.6 mL of nitrogen collected
would be produced from the complete reaction of 25 mL of 0.833 mol/L 
with excess 
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