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 (Na 2 SiO 3 ), for example, reacts as follows: Na 2 SiO 3 (s) + 8 HF(aq) → H 2 SiF 6 (aq) + 2 NaF(aq) + 3 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?

Welcome everyone. Here's our next question. Hydrofluoric acid HF aqueous cannot be stored in glass bottles because compounds called silicates in the glass are attacked by the HF sodium silicate N A two SI 03. For an example, rea as followed N A two SI 03 solid plus eight HF aqueous becomes H two si F six aqueous plus two NAF aqueous plus three water liquid. How many grams of NAF form when 0.630 moles of HF reacts with excess N A two SI 03 A 10.4 g. B 5.54 g, C 6.61 g or D 1.39 g. So let's think about our notes and what we're trying to find out. So we're given the mass of sodium fluoride or excuse me, we're looking for is our unknown, the mass of sodium fluoride that's produced. We're given the number of moles of our reactant. This is our limiting reagent since our other reagent is present in excess. So we want to convert that to moles of our sodium fluoride product and then using molar mass, 2 g of sodium fluoride. So pretty straightforward here. So we do need to double check that our equation is balanced before we make that conversion. So N I two N A two, excuse me, SI 03 plus eight HF we see these coefficients. So the assumption is this is balanced but always wise just to double check, make sure there's nothing wrong. H two SI F six. It's our first product plus two NAF plus 3 H2O. So we see we've got two sodiums on both sides. We've got one silicon on both sides. Oxygen, we've got three on the left, three on the right, eight hydrogens on the left. And then we see two plus six on the right eight fluorine. And then on the right, we have six plus two fluorine. So our equation is balanced, we'll come free, free to use that. So we start with our moles. So 0.630 moles of hydrogen fluoride. Then we use a conversion factor to go to moles of sodium fluoride. So we see from our balanced equation, we'll put moles of hydrogen fluoride on the denominator. So they'll cancel out. We have eight moles HF in the denominator. And then in the numerator, we have two moles of sodium fluoride. So every eight moles of our hydrogen fluoride that we consume, we produce two moles of sodium fluoride. And then we just need the molar mass of sodium fluoride to calculate that. So we'll just do a quick calculation of the molar mass of sodium fluoride, we've just got one atom of each. So the atomic mass of sodium is 22.99 g per mole plus the atomic mass of fluorine, which is 19.00 g per mole. Giving us a total molar mass of 41.99 g per mole. So let's make a conversion factor with that. So we've gone to moles of sodium fluoride. We multiply that. So we know one mole of sodium fluoride goes on the bottom of that conversion factor, the denominator and the numerator, then we have 41.99 g sodium fluoride. So our moles of HF cancel out our moles of sodium fluoride. Cancel out, we're left with grams of sodium fluoride, which is indeed what were looking for. So we'll plug those numbers into our calculator. So we have 0.630 multiplied by two, then divided by eight, multiplied by 41.99. Our quantity given to us has three sig figs. So our answer must also have three. So the answer will lead to 6.61 g of sodium fluoride. So there's my answer. When I go over to my answer choices, I see that matches choice C So again, when I have 0.630 moles of hydrogen fluoride reacting with excess N A two SI 03, we get choice C 6.61 g of sodium fluoride. See you in the next video.

Ch.3 - Chemical Reactions and Reaction Stoichiometry

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Brown 14th Edition

Ch.3 - Chemical Reactions and Reaction Stoichiometry

Problem 61

Chapter 3, Problem 61

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 (Na₂SiO₃), for example, reacts as follows: Na₂SiO₃(s) + 8 HF(aq) → H₂SiF₆(aq) + 2 NaF(aq) + 3 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 Na₂SiO₃can react with 0.800 g of HF?

Verified step by step guidance

Identify the balanced chemical equation: \( \text{Na}_2\text{SiO}_3(s) + 8 \text{HF}(aq) \rightarrow \text{H}_2\text{SiF}_6(aq) + 2 \text{NaF}(aq) + 3 \text{H}_2\text{O}(l) \).

Determine the molar ratio between HF and NaF from the balanced equation. The ratio is 8 moles of HF to 2 moles of NaF.

Calculate the moles of NaF produced from 0.500 moles of HF using the molar ratio: \( \frac{2 \text{ moles NaF}}{8 \text{ moles HF}} \times 0.500 \text{ moles HF} \).

Convert the moles of NaF to grams using the molar mass of NaF. The molar mass of NaF is approximately 41.99 g/mol.

Multiply the moles of NaF by its molar mass to find the mass in grams: \( \text{moles of NaF} \times 41.99 \text{ g/mol} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on balanced equations. It allows chemists to determine the amount of substances consumed and produced in a reaction. In this case, understanding the stoichiometric coefficients from the balanced equation is essential to find out how many grams of NaF are produced from a given amount of HF.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is crucial for converting between moles and grams in stoichiometric calculations. For NaF, knowing its molar mass allows us to calculate the mass of NaF produced from the moles derived from the reaction with HF.

Limiting Reactant

The limiting reactant is the substance that is completely consumed first in a chemical reaction, thus determining the maximum amount of product formed. In this scenario, since HF is provided in a specific amount and Na2SiO3 is in excess, HF will limit the formation of NaF. Identifying the limiting reactant is key to accurately calculating the yield of NaF.

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Textbook Question

The reaction between potassium superoxide, KO2, and CO2, 4 KO2 + 2 CO2¡2K2CO3 + 3 O2 is used as a source of O2 and absorber of CO2 in selfcontained breathing equipment used by rescue workers. (a) How many moles of O2 are produced when 0.400 mol of KO2 reacts in this fashion?

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Textbook Question

The reaction between potassium superoxide, KO2, and CO2, 4 KO2 + 2 CO2¡2K2CO3 + 3 O2 is used as a source of O2 and absorber of CO2 in selfcontained breathing equipment used by rescue workers. (b) How many grams of KO2 are needed to form 7.50 g of O2?

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Textbook Question

Several brands of antacids use Al1OH23 to react with stomach acid, which contains primarily HCl: Al(OH)₃(s) + HCl(aq) → AlCl₃(aq) + H₂O(l) (b) Calculate the number of grams of HCl that can react with 0.500 g of Al(OH)₃. (c) Calculate the number of grams of AlCl₃ and the number of grams of H₂Oformed when 0.500 g of Al(OH)₃ reacts.

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