Calculations Of Moles And Grams When Hydrofluoric Acid Reacts With Sodium Silicate

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Calculations Of Moles And Grams When Hydrofluoric Acid Reacts With Sodium Silicate

The ease with which hydrofluoric acid (HF) reacts with silicate compounds is a reason hydrofluoric acid is not stored in glass bottles (glass contain silicate compounds).

Given the following chemical reaction between hydrofluoric acid and sodium silicate:

Na_{2}SiO_{3}(s) + 8HF(aq) ----> H_{2}SiF_{6}(aq) + 2NaF(aq) + H_{2}O(l).

(a) How many moles of HF are needed to react with 0.300 mol of Na_{2}SiO_{3}?

(b) How many grams of NaF form when 0.500 mol of HF reacts with excess Na_{2}SiO_{3}?

(c) How many grams of Na_{2}SiO_{3} can react with 0.800 g of HF?

**The strings**:
S_{7}P_{6}A_{66} (interaction - chemical).

S_{7}P_{1}A_{15} (containership - mass).
**The math**:

Although we are calculating , mass, grams and moles, the chemical reaction is the reason for the calculation. So Pj Problem of Interest is of type *interaction* (chemical).

**(a)** 1 mole of Na_{2}SiO_{3} reacts with 8 moles of HF

So, 0.300 moles of Na_{2}SiO_{3} will react with 0.300(8) HF = 2.4 mols of HF.
**(b)** 8 moles of HF forms 2 moles of NaF

So, 0.500 moles of HF forms 2(o.500)/8 = 0.125 mol of NaF

Molar mass of NaF = 23 + 19 = 42

So, 0.125 mol of NaF has mass 0.125(42) = 5.25 g

So, 5.25 g forms when 0.500 mol of HF reacts with excess Na_{2}SiO_{3}.
**(c)** Molar mass of HF = 1 + 19 = 20

0.800 g of HF = 0.800/20 mol of HF = 0.04 mol of HF

8 moles of HF reacts with 1 mole Na_{2}SiO_{3}

So, 0.04 mol of HF will react with 0.04(8) mol of Na_{2}SiO_{3} = 0.005 mol.

Molar mass of Na_{2}SiO_{3} = 46 + 48 + 28 = 122.

So, 0.005 mol of Na_{2}SiO_{3} = 0.005(122) g of Na_{2}SiO_{3} = 0.610 g.

So, 0.610 g of Na_{2}SiO_{3} can react with 0.800 g of HF.

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