Answer to Question #24957 in Inorganic Chemistry for jordan

Question

How many grams of sodium silicate can be reacted with 8.2 g of hydrofluoric acid?

Answer 1

Thechemical equation for this reaction is

Na₂SiO₃ + 8HF = H₂SiF₆ + 2NaF + 3H₂O

M_W(HF) = M_W(H) + M_W(F) = 1.0 + 19.0 = 20.0 g/mol
M_W(Na₂SiO₃) = 2 x M_W(Na) + M_W(Si) + 3 x M_W(O) = 2 x 23.0 + 28.1 + 3 x 16.0 =122.1 g/mol

The amount of hydrofluoric acid:

n(HF) = m(HF)/M_W(HF) = 8.2/20.0 = 0.41 moles

The amount of sodium silicate needed is

n(Na₂SiO₃) = n(HF)/8 = 0.051 moles

The weight of sodium silicate

m(Na₂SiO₃) = n(Na₂SiO₃) x M_W(Na₂SiO₃) = 0.051 x 122.1 = 6.23 g

Answer: m(Na₂SiO₃) = 6.23 grams