Glass is a mixture of several compounds, but a major constituent of most glass is calcium silicate, \(\mathrm{CaSiO}_{3}\). Glass can be etched by treatment with hydrofluoric acid; HF attacks the calcium silicate of the glass, producing gaseous and water-soluble products (which can be removed by washing the glass). For example, the volumetric glassware in chemistry laboratories is often graduated by using this process. Balance the following equation for the reaction of hydrofluoric acid with calcium silicate. $$\mathrm{CaSiO}_{3}(s)+\mathrm{HF}(a q) \longrightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$

Short Answer

Expert verified
The balanced chemical equation for the reaction of hydrofluoric acid with calcium silicate is: $$\mathrm{CaSiO}_{3}(s)+6\mathrm{HF}(aq) \rightarrow \mathrm{CaF}_{2}(aq)+\mathrm{SiF}_{4}(g)+3\mathrm{H}_{2} \mathrm{O}(l)$$

Step by step solution

01

Identify the elements in the reaction and count their atoms initially

First, let's identify the elements involved in the reaction. We have Calcium (Ca), Silicon (Si), Oxygen (O), Hydrogen (H), and Fluorine (F). Now, count the number of atoms for each element on both sides of the equation: Reactants: - Ca: 1 - Si: 1 - O: 3 - H: 1 - F: 1 Products: - Ca: 1 - Si: 1 - O: 1 - H: 2 - F: 6
02

Balance the equation one element at a time

To balance the equation, we will start adjusting the coefficients for the compounds containing the least common elements and work our way through to the most common elements (usually H and O). 1. Balance F: Since there are 6 F atoms in the products and only 1 F atom in the reactants, add a coefficient of 6 to the HF reactant: $$\mathrm{CaSiO}_{3}(s)+6\mathrm{HF}(a q) \longrightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Now there are 6 H atoms and 6 F atoms in the reactants. 2. Balance O: There are 3 O atoms in the reactants and only 1 O atom in the products. Add a coefficient of 3 to the H2O product to balance the O atoms: $$\mathrm{CaSiO}_{3}(s)+6\mathrm{HF}(a q) \longrightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+3\mathrm{H}_{2} \mathrm{O}(l)$$ Now there are 6 H atoms and 3 O atoms in both the reactants and products. 3. Check the balance for Ca and Si: There is no need to adjust the coefficients for Ca and Si, as there is already 1 Ca and 1 Si atom on both sides of the equation. At this point, the equation is balanced. The balanced chemical equation for the reaction of hydrofluoric acid with calcium silicate is: $$\mathrm{CaSiO}_{3}(s)+6\mathrm{HF}(aq) \rightarrow \mathrm{CaF}_{2}(aq)+\mathrm{SiF}_{4}(g)+3\mathrm{H}_{2} \mathrm{O}(l)$$

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Most popular questions from this chapter

Anabolic steroids are performance enhancement drugs whose use has been banned from most major sporting activities. One anabolic steroid is fluoxymesterone \(\left(\mathrm{C}_{20} \mathrm{H}_{29} \mathrm{FO}_{3}\right) .\) Calculate the percent composition by mass of fluoxymesterone.

In the production of printed circuit boards for the electronics industry, a 0.60-mm layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q)$$ A plant needs to manufacture 10,000 printed circuit boards, each \(8.0 \times 16.0 \mathrm{cm}\) in area. An average of \(80 . \%\) of the copper is removed from each board (density of copper \(=8.96 \mathrm{g} / \mathrm{cm}^{3}\)). What masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) are needed to do this? Assume \(100 \%\) yield.

A given sample of a xenon fluoride compound contains molecules of the type \(\mathrm{XeF}_{n},\) where \(n\) is some whole number. Given that \(9.03 \times 10^{20}\) molecules of \(\mathrm{XeF}_{n}\) weigh \(0.368 \mathrm{g},\) determine the value for \(n\) in the formula.

In using a mass spectrometer, a chemist sees a peak at a mass of \(30.0106 .\) Of the choices \(^{12} \mathrm{C}_{2}\) \(^{1} \mathrm{H}_{6},\) \(^{12} \mathrm{C}\) \(^{1} \mathrm{H}_{2}\) \(^{16} \mathrm{O},\) and \(^{14} \mathrm{N}^{16} \mathrm{O},\) which is responsible for this peak? Pertinent masses are \(^{1} \mathrm{H}\), \(1.007825 ;^{16} \mathrm{O}, 15.994915 ;\) and \(^{14} \mathrm{N}, 14.003074.\)

Natural rubidium has the average mass of 85.4678 u and is composed of isotopes \(^{85} \mathrm{Rb}\) (mass \(=84.9117 \mathrm{u}\) ) and \(^{87} \mathrm{Rb}\). The ratio of atoms \(^{85} \mathrm{Rb} /^{87} \mathrm{Rb}\) in natural rubidium is \(2.591 .\) Calculate the mass of \(^{87} \mathrm{Rb}\).

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