1. According to the equation, what mass of hydrogen fluoride is necessary to produce 2.3 g of sodium fluoride? HF + NaNO3 -> HNO3 + NaF

1.09 grams

Explanation:

According to the following chemical equation:

HF + NaNO₃ -> HNO₃ + NaF

1 mol of hydrogen fluoride (HF) produces 1 mol of sodium fluoride (NaF). Thus, we first convert from mol to grams by using the molar mass (MM) of each compound:

MM(HF)= (1 g/mol x 1 H) + (19 g/mol x 1 F) = 20 g/mol HF

1 mol HF x 19.9 g/mol HF = 20 g

MM(NaF) = (23 g/mol x 1 Na) + (19 g/mol x 1 F) = 42 g/mol NaF

1 mol NaF x 42 g/mol NaF = 42 g

Thus, from 20 g of HF are produced 42 g of NaF  ⇒ 20 g HF/42 g NaF. We multiply this stoichiometric ratio by the mass of NaF produced to calculate the required mass of HF:

20 g HF/42 g NaF x 2.3 g NaF = 1.09 g HF

Therefore, 1.09 grams of HF are necessary to produce 2.3 g of NaF.

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