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Answer on Inorganic Chemistry Question for jordan

Question #24957
How many grams of sodium silicate can be reacted with 8.2 g of hydrofluoric acid?
Expert's answer
Thechemical equation for this reaction is

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

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

The amount of hydrofluoric acid:

n(HF) = m(HF)/MW(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 MW(Na2SiO3) = 0.051 x 122.1 = 6.23 g

Answer: m(Na2SiO3) = 6.23 grams

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