Hydrofluoric acid, \(\mathrm{HF}(a q),\) cannot be stored in glass bottles because compounds called silicates in the glass are attacked by the \(\mathrm{HF}(a q)\). Sodium silicate \(\left(\mathrm{Na}_{2} \mathrm{SiO}_{3}\right)\), for example, reacts as follows: $$ \begin{aligned} \mathrm{Na}_{2} \mathrm{SiO}_{3}(s)+8 \mathrm{HF}(a q) & \longrightarrow \\ \mathrm{H}_{2} \mathrm{SiF}_{6}(a q) &+2 \mathrm{NaF}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ (a) How many moles of HF are needed to react with 0.300 \(\mathrm{mol}\) of \(\mathrm{Na}_{2} \mathrm{SiO}_{3} ?\) (b) How many grams of NaF form when \(0.500 \mathrm{~mol}\) of HF reacts with excess \(\mathrm{Na}_{2} \mathrm{SiO}_{3} ?\) (c) How many grams of \(\mathrm{Na}_{2} \mathrm{SiO}_{3}\) can react with \(0.800 \mathrm{~g}\) of HF?

Step by step solution

01

Determine the mole ratio of HF to Na2SiO3

From the balanced equation: Na2SiO3(s) + 8HF(aq) -> H2SiF6(aq) + 2NaF(aq) + 3H2O(l), it can be seen that the mole ratio of HF to Na2SiO3 is 8:1.

02

Calculate moles of HF required

Using the mole ratio of HF to Na2SiO3, we can calculate the moles of HF needed to react with 0.300 mol of Na2SiO3. Moles of HF = (0.300 mol Na2SiO3) × (8 mol HF / 1 mol Na2SiO3) = 2.4 mol HF So, 2.4 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?

03

Determine the mole ratio of NaF to HF

From the balanced equation: Na2SiO3(s) + 8HF(aq) -> H2SiF6(aq) + 2NaF(aq) + 3H2O(l), it can be seen that the mole ratio of NaF to HF is 2:8 or 1:4.

04

Calculate moles of NaF formed

Using the mole ratio of NaF to HF, we can calculate the moles of NaF formed when 0.500 mol of HF reacts with excess Na2SiO3. Moles of NaF = (0.500 mol HF) × (1 mol NaF / 4 mol HF) = 0.125 mol NaF

05

Calculate grams of NaF formed

To calculate the grams of NaF formed, we need to find the molar mass of NaF. Na has a molar mass of 22.99 g/mol and F has a molar mass of 19.00 g/mol. Molar mass of NaF = 22.99 + 19.00 = 41.99 g/mol Now, we can multiply the moles of NaF by its molar mass to get the grams of NaF. Grams of NaF = (0.125 mol NaF) × (41.99 g/mol) = 5.25 g NaF So, 5.25 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?

06

Convert grams of HF to moles

First, we need to find the molar mass of HF. H has a molar mass of 1.01 g/mol and F has a molar mass of 19.00 g/mol. Molar mass of HF = 1.01 + 19.00 = 20.01 g/mol Now, we can convert the grams of HF to moles. Moles of HF = (0.800 g HF) / (20.01 g/mol) = 0.040 moles HF

07

Calculate moles of Na2SiO3

Using the mole ratio of Na2SiO3 to HF, we can calculate the moles of Na2SiO3 that can react with 0.040 moles of HF. Moles of Na2SiO3 = (0.040 mol HF) × (1 mol Na2SiO3 / 8 mol HF) = 0.005 mol Na2SiO3

08

Calculate grams of Na2SiO3

To calculate the grams of Na2SiO3, we need to find the molar mass of Na2SiO3. Na has a molar mass of 22.99 g/mol, Si has a molar mass of 28.09 g/mol and O has a molar mass of 16.00 g/mol. Molar mass of Na2SiO3 = (2 × 22.99) + 28.09 + (3 × 16.00) = 122.07 g/mol Now, we can multiply the moles of Na2SiO3 by its molar mass to get the grams of Na2SiO3. Grams of Na2SiO3 = (0.005 mol Na2SiO3) × (122.07 g/mol) = 0.61 g Na2SiO3 So, 0.61 grams of Na2SiO3 can react with 0.800 g of HF.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!