Hydrofluoric acid solutions cannot be stored in glass containers because hf reacts readily with silica dioxide in glass to produce hexafluorosilicic acid (h2sif6).
sio2(s) + 6 hf(aq) h2sif6(aq) + 2 h2o(l)
21.0 g sio2 and 70.5 g of hf react to yield 45.8 g h2sif6.

(a) what is the limiting reactant?

hf
or
sio2

(b) what is the mass of the excess reactant?

g

(c) what is the theoretical yield of h2sif6?

g

(d) what is the percent yield?

%

A) Limiting reactant

You need the molar ratios (from the balanced chemical equation) and the molar masses of each compound (from the atomic masses)

a) Molar ratios:

6 mol HF : 1 mol SiO2 : 1 mol H2SiF6

2) Molar masses:

Atomic masses:
H: 1 g/mol
F: 19 g/mol
Si: 28 g/mol
O: 16g/mol

=>
HF:1g/mol + 19 g/mol = 20 g/mol
SiO2: 28g/mol + 2*16g/mol = 60 g/mol
H2SiF6: 2*1g/mol + 28g/mol + 6*19g/mol = 144g/mol

3) convert data in grams to moles

21.0 g SiO2 / 60 g/mol = 0.35 mol SiO2

70.5 g HF /  20 g/mol = 3.525 mol HF

4) Use the theorical ratios to deduce which is in excess and which is the limiting reactant.

6 mol HF / 1mol SiO2   < 3.525 mol HF / 0.35 mol SiO2 ≈ 10

=> There is more HF than the needed to react with 0.35mol of SiO2 =>

SiO2 is the limiting reactant (HF is in excess)

b) Mass of excess reactant.

1) Calculate how many grams reacted, which requires to calculate first the number of moles that reacted

0.35 mol SiO2 * 6 mol HF / 1 mol SiO2 = 2.1 mol of HF

2.1 mol HF * 20 g/mol = 42 gram of HF

2) Subtract the quantity that reacted from the original quantity:

70.5 g - 42 g = 28.5 g of HF in excess

c) Theoretical yield of H2SiF6

1 mol of SiO2 ; 1 mol of H2SiF6 => 0.35 mol SiO2 : 0.35 mol H2SiF6

Convert those moles to grams: 0.35 mol * 144 g/mol = 50.4 grams

d) % yield

% yield = actual yield / theoretical yield * 100 = 45.8 / 50.4 * 100 = 90.87%

Given data,temperature, t = 230°c = 503 kconsider the chemical reaction,ch₃oh(l) + 1/2 o₂(g)  → ch₂o(g) + h₂o(l)calculate the enthalpy change for the reaction: δh⁰rxn =  σnδh⁰f(products) -  σmδh⁰f(reactants)substitute the values from the table: δh⁰rxn = { -115.9 + (-285.8)} - {0+(-238.7)}              = -163.0 kjcalculate the entropy change for the reaction.δs⁰rxn = (219 + 70) - (127.2 + 1/2 x 205.2) = 59.2 j/kconsider the expression for free energy: δg⁰rxn =  δh⁰rxn - t  δs⁰rxnsubstitute the values in the above equationδg⁰rxn = -163 x 10³ - (503 x 59.2)              = -192777.6 j              = -192.777 kj