If 32 ml of 7.0 m h2so4 was spilled, what is the minimum mass of nahco3 that must be added to the spill to neutralize the acid?

The complete balanced reaction of this neutralization reaction is:

 H2SO4  +  2NaHCO3  -->  2CO2  +  2H2O  +  Na2SO4

Then we calculate the moles of H2SO4 that was spilled:

 moles H2SO4 = 7 mole/L * 0.032 L = 0.224 mole

From the reaction, we see that 2 moles of NaHCO3 is required for every mole of H2SO4, hence:

 moles NaHCO3 = 0.224 * 2 = 0.448 mole

The molar mass of NaHCO3 is 84 g/mol. Hence the mass is:

mass NaHCO3 = 0.448 * 84

mass NaHCO3 = 37.632 grams

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.

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