Question

Hydrofluoric acid, HF(aq), cannot be stored in glass bottles because compounds called silicates in the glass are attacked by the HF(aq). Sodium silicate ( $ N{a_2}Si{O_3} $ ), for example, reacts as follows: $ N{a_2}Si{O_{3(s)}} + 8H{F_{(aq)}} \to {H_2}Si{F_{6(aq)}} + 2Na{F_{(aq)}} + 3{H_2}{O_{(l)}} $ ?
A) How many moles of HF are needed to react with 0.260 mol of $ N{a_2}Si{O_3} $ ?
B) How many grams of NaF form when 0.600 mol of HF reacts with excess $ N{a_2}Si{O_3} $ ?
C) How many grams of $ N{a_2}Si{O_3} $ can react with 0.900 g of HF?

Answer 1

Hint: To solve this problem we will consider the mole ratio of the reactants and products. Mole ratio is the smallest whole number ratio of the reactants and products. Each reactant/product ratio is separated by ‘:’ and that of reactant and product is separated by ‘: :’.

Complete Step By Step Answer:
The balanced chemical equation of the reaction is: $ N{a_2}Si{O_{3(s)}} + 8H{F_{(aq)}} \to {H_2}Si{F_{6(aq)}} + 2Na{F_{(aq)}} + 3{H_2}{O_{(l)}} $
The mole ratio of this reaction is: $ 1:8 = 1:2:3 $
We’ll solve the subparts one by one.
A) How many moles of HF are needed to react with 0.260 mol of $ N{a_2}Si{O_3} $ ?
To solve this we will need the ratio formula. The ration formula of the given reaction is: $ 1mol{\text{ }}N{a_2}Si{O_3}:8mol{\text{ }}HF = 1mol{\text{ }}{H_2}Si{F_6}:2mol{\text{ }}NaF:3{\text{ }}mol{\text{ }}{H_2}O $
Solving the mole ratio we get (given that we have 0.26 mol of $ N{a_2}Si{O_3} $ ):
 $ (0.26 \times 1):(0.26 \times 8) = (0.26 \times 1):(0.26 \times 2):(0.26 \times 3) $
 $ 0.26mol{\text{ }}N{a_2}Si{O_3}:2.08mol{\text{ }}HF = 0.26mol{\text{ }}{H_2}Si{F_6}:0.52mol{\text{ }}NaF:0.78mol{\text{ }}{H_2}O $
From the above relation we can conclude that 0.26 mol of $ N{a_2}Si{O_3} $ will need 2.08 mol of HF
B) How many grams of NaF forms when 0.600 mol of HF reacts with excess $ N{a_2}Si{O_3} $ ?
We have found out the ratio formula. We know that 1 mol $ N{a_2}Si{O_3} $ will require 8 mol of HF. Therefore 0.600 mol of HF will require $ \dfrac{{0.600}}{8}mol $ $ N{a_2}Si{O_3} $ . using the unitary method we will use the variable x such that $ \dfrac{{0.600}}{8} \times x({\text{variable}}) $
Using this in the above ratio formula we get:
 $ \dfrac{{0.6}}{8} \times 1mol{\text{ }}N{a_2}Si{O_3}:\dfrac{{0.6}}{8} \times 8mol{\text{ }}HF = \dfrac{{0.6}}{8} \times 1mol{\text{ }}{H_2}Si{F_6}:\dfrac{{0.6}}{8} \times 2mol{\text{ }}NaF:\dfrac{{0.6}}{8} \times 3{\text{ }}mol{\text{ }}{H_2}O $
 $ 0.075mol{\text{ }}N{a_2}Si{O_3}:0.6mol{\text{ }}HF = 0.075mol{\text{ }}{H_2}Si{F_6}:0.15mol{\text{ }}NaF:0.225mol{\text{ }}{H_2}O $
Therefore, we can conclude that 0.15 moles of NaF is formed. The molar mass of NaF is 41.9887g/mol. Therefore, the mass of NaF is given as: $ moles \times molar{\text{ }}Mass $
The mass of NaF formed from 0.6 moles of HF $ = 41.98817 \times 0.15 = 6.298g $
C) How many grams of $ N{a_2}Si{O_3} $ can react with 0.900 g of HF?
The no. of moles of HF in 0.900g of HF is given by the formula: $ moles = \dfrac{{mass}}{{molar{\text{ }}mass}} $
Therefore, no. of moles of HF $ = \dfrac{{0.9}}{{20.01}} = 0.045mol $
The ratio of $ N{a_2}Si{O_3} $ to HF is given by the ratio formula as: $ 1:8 $
Using the unitary method again we can write the ratio formula as:
 $ \dfrac{{0.045}}{8} \times 1mol{\text{ }}N{a_2}Si{O_3}:0.045mol{\text{ }}HF $
 $ 0.00563mol{\text{ }}N{a_2}Si{O_3}:0.045mol{\text{ }}HF $
Mass of $ N{a_2}Si{O_3} $ = $ moles \times molar{\text{ }}Mass $
The molar mass of $ N{a_2}Si{O_3} $ = 122.06 g/mol. Mass of $ N{a_2}Si{O_3} $ = $ 0.00563 \times 122.06 = 0.6871g $ .

Note:
Remember that while finding no. of moles take the mass in grams and the molar mass in g/mol. If the mass unit is different (kg,mg etc) either convert it into g or use the molar mass also in that unit itself. In this question we have only found out the formula ratio, and found out the mass of reactants accordingly.