The Solvay process is used for the manufacture baking soda, NaHCO 3 . In the pro

Question

The Solvay process is used for the manufacture baking soda, NaHCO3. In the process, CO2, NH3, H20, and NaCL react to produce baking soda. If 25.0 L CO2 and 20.0 L NH3 react at STP, with excess water and sodium chloride, what is the limiting reactant? Calculate the mass of baking soda produced.

CO2(g) + NH3(g) + H2O(l) + NaCL (s) --> NaHCO3(s) + NH4CL(aq)

If someone can please help me I'm totally stuck. If you decide to answer this by posting a picture please have the picture be non blurry so I can understand. Chemistry is NOT my thing!!!

Explanation / Answer

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CO2(g) + NH3(g) + H2O(l) + NaCL (s) --> NaHCO3(s) + NH4CL(aq

At STP 22.4 L of any gas weighs 1 mole

So , 25.0 L of CO2 = (1mole Co2 / 22.4 L Co2) * 25.0 L CO2 = 1.12 mole CO2

So , 20.0 L of NH3   = (1mole NH3 / 22.4 L NH3 ) * 20.0 L NH3 = 0.893 mole NH3

As per reaction 1 mole CO2 needs 1 mole NH3 to react

Hence 1.12 mole CO2 will also need 1.12 moles of NH3 but we have only 0.893 moles NH3 which are less than required ,

so NH3 is limiting reactant and used up completely to form 0.893 moles NaHCO3

moles of NaHCO3 formed = 0.893 moles

grams of NaHCo3 = moles NaHCO3 * molar mass NaHCO3

= 0.893 * 84.00

= 75.012 g (answer)

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