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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:
Na2SiO3(s) + 8HF(aq) ----> H2SiF6(aq) + 2NaF(aq) + H2O(l).
(a) How many moles of HF are needed to react with 0.300 mol of Na2SiO3?
(b) How many grams of NaF form when 0.500 mol of HF reacts with excess Na2SiO3?
(c) How many grams of Na2SiO3 can react with 0.800 g of HF?
The strings:
S7P6A66 (interaction - chemical).
S7P1A15 (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 Na2SiO3 reacts with 8 moles of HF
So, 0.300 moles of Na2SiO3 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 Na2SiO3.
(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 Na2SiO3
So, 0.04 mol of HF will react with 0.04(8) mol of Na2SiO3 = 0.005 mol.
Molar mass of Na2SiO3 = 46 + 48 + 28 = 122.
So, 0.005 mol of Na2SiO3 = 0.005(122) g of Na2SiO3 = 0.610 g.
So, 0.610 g of Na2SiO3 can react with 0.800 g of HF.
The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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