Hydrofluoric acid, 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: $$ \mathrm{Na}_{2} \mathrm{SiO}_{3}(s)+8 \mathrm{HF}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{SiF}_{6}(a q)+2 \mathrm{NaF}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) How many moles of HF are needed to react with 0.300 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

Identify the balanced chemical equation

The chemical equation given is: Na2SiO3(s) + 8HF(aq) -> H2SiF6(aq) + 2NaF(aq) + 3H2O(l)

02

Use stoichiometry to determine moles of HF

We will use the stoichiometry of the balanced equation to determine the moles of HF needed to react with 0.300 mol of Na2SiO3: 1 mol Na2SiO3 needs 8 mol HF So, 0.300 mol Na2SiO3 will need: 0.300 mol Na2SiO3 × (8 mol HF / 1 mol Na2SiO3) = 2.40 mol HF #b) Calculate grams of NaF when 0.500 mol HF reacts with excess Na2SiO3#

03

Use stoichiometry to determine moles of NaF

We are given that 0.500 mol HF reacts with excess Na2SiO3. We will use the stoichiometry of the balanced equation to determine the moles of NaF formed: 8 mol HF produces 2 mol NaF So, 0.500 mol HF will produce: 0.500 mol HF × (2 mol NaF / 8 mol HF) = 0.125 mol NaF

04

Convert moles of NaF to grams

Now, we will use the molar mass of NaF (22.99 g/mol for Na + 18.998 g/mol for F = 41.988 g/mol) to convert moles of NaF to grams: 0.125 mol NaF × (41.988 g/mol) = 5.25 g NaF #c) Calculate grams of Na2SiO3 that can react with 0.800 g HF#

05

Convert grams of HF to moles

We will use the molar mass of HF (1.008 g/mol for H + 18.998 g/mol for F = 20.006 g/mol) to convert the given mass of HF (0.800 g) to moles: 0.800 g HF × (1 mol HF / 20.006 g/mol) = 0.0400 mol HF

06

Use stoichiometry to determine moles of Na2SiO3

We will use the stoichiometry of the balanced equation to determine the moles of Na2SiO3 that can react with 0.0400 mol HF: 8 mol HF reacts with 1 mol Na2SiO3 So, 0.0400 mol HF will react with: 0.0400 mol HF × (1 mol Na2SiO3 / 8 mol HF) = 0.00500 mol Na2SiO3

07

Convert moles of Na2SiO3 to grams

Now, we will use the molar mass of Na2SiO3 (2 × 22.99 g/mol for Na + 28.09 g/mol for Si + 3 × 15.999 g/mol for O = 122.07 g/mol) to convert moles of Na2SiO3 to grams: 0.00500 mol Na2SiO3 × (122.07 g/mol) = 0.610 g Na2SiO3 To sum up the answers: a) 2.40 mol HF are needed to react with 0.300 mol of Na2SiO3 b) 5.25 grams of NaF form when 0.500 mol of HF reacts with excess Na2SiO3 c) 0.610 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!