22. a flask containing 450 ml of 0.50 m h2so4 was accidentally knocked to the floor. how many grams of nahco, do you need to put on the spill to neutralize the acid according to the following equation: h2so4(aq)+2 nahcos(aq) na, so(aq) +2 h20()+2 co2(g) d) 38 g a) 2.3 g b) 9.5 g c) 19 g

The correct answer is Option D.

Explanation:

To calculate the moles of a solute, we use the equation:

We are given:

Volume of sulfuric acid = 450mL = 0.45 L      (Conversion factor: 1 L = 1000 mL)

Molarity of the solution = 0.5 moles/ L

Putting values in above equation, we get:

For the given chemical reaction:

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of sodium bicarbonate.

So, 0.225 moles of sulfuric acid will react with = of sodium bicarbonate

To calculate the mass of sodium bicarbonate, we use the equation:

Moles of sodium bicarbonate = 0.45 moles

Molar mass of sodium bicarbonate = 84.007 g/mol

Putting values in above equation, we get:

Hence, the correct answer is Option D.

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g