Question

The reaction shown below is a way to generate sodium nitrate for fabric dye production and as a meat preservative. Carbon dioxide is simultaneously produced. Calculate a) the mass of CO2 generated and b) the concentration of Na+ ions in solution when the reaction has gone to completion, given the following starting materials: 24,500 ml of 2.50 M Na2CO3, 3.6 kg NO and 20.0 kg of O2. Assume 100% yield. 2Na2CO3(aq) + 4NO(g) + O2(g) --> 4NaNO2(aq) + 2CO2(g)

Answer 1

From the question,

24500ml of 2.5 M Na₂CO₃contains 61.25 moles of Na₂CO_3. from the molarity formula which 2.5 = (no. of molesX1000)/24500

3.6 Kg of NO contains 120 moles of NO

20 kg of O₂ contains 625 moles of O₂

according to the chemical equation for the reaction, 2Na₂CO₃(aq) + 4NO(g) + O₂(g) --> 4NaNO₂(aq) + 2CO2(g), which can be written as Na₂CO₃(aq) + 2NO(g) + 1/2O₂(g) --> 2NaNO₂(aq) + CO₂(g)

each mole of Na₂CO₃ should react with 2 moles of NO, hence 61.25 moles of Na₂CO₃ needs 122.5 moles of NO. But, we have only 120 moles of NO, thus NO is limiting reagent in this reaction. So, according to chemical reaction 2 moles of NO will produce 1 mole of CO₂. After complete reaction, only 60 moles of CO₂ will be produced which is equivalent to 2.64 Kg of CO₂ by mass.

Similarly, 120 moles of NO will generate 120 moles of NaNO₂ after completion of reaction. Thus 120 moles of Na⁺ ions present in 24500 ml, hence concentration is 4.89 M