Hydrofluoric acid, hf(aq), cannot be stored in glass bottles because compounds called silicates in the glass are attacked by the hf(aq). sodium silicate (na2sio3), for example, reacts as follows: na2sio3(s)+8hf(aq)→h2sif6(aq)+2naf( aq)+3h2o(l)? a)how many moles of hf are needed to react with 0.260 mol of na2sio3? b) how many grams of naf form when 0.600 mol of hf reacts with excess na2sio3? c)how many grams of na2sio3 can react with 0.900 g of hf?

Explanation:

1. Moles of HF

We know we will need an equation with moles, so let’s gather all the information in one place.  

         Na₂SiO₃ + 8HF ⟶ H₂SiF₆ + 2NaF + 3H₂O  

n/mol: 0.260  

The molar ratio is 8 mol HF:1 mol Na₂SiO₃  

2. Mass of NaF  

M_r:   122.06  

         Na₂SiO₃ + 8HF ⟶ H₂SiF₆ + 2NaF + 3H₂O  

m/g:                  0.600  

The molar ratio is 2 mol NaF:8 mol HF  

3. Mass of Na₂SiO₃  

M_r:  122.06    20.01  

        Na₂SiO₃ + 8HF ⟶ H₂SiF₆ + 2NaF + 3H₂O  

m/g:                  0.900  

The molar ratio is 1 mol Na₂SiO₃:8 mol HF  

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

it means there are 3 atoms of oxygen