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Chapter 21: Problem 42

Write a chemical equation to represent (a) the reaction of potassium cyanide solution with silver nitrate solution; (b) the combustion of \(\mathrm{Si}_{3} \mathrm{H}_{8}\) in an excess of oxygen; (c) the reaction of dinitrogen with calcium carbide to give calcium cyanamide (CaNCN).

Short Answer

Expert verified
(a) KCN(aq) + AgNO3(aq) → AgCN(s) + KNO3(aq) \n(b) 3Si3H8(g) + 16O2(g) → 9SiO2(s) + 12H2O(g) \n(c) N2(g) + 3CaC2(s) → 3CaNCN(s) + C(s)

Step by step solution

01

(a) Reaction of potassium cyanide with silver nitrate

To write the chemical equation, identify the reactants and products. For this reaction, potassium cyanide (KCN) and silver nitrate (AgNO3) are reactants. When these two react, they undergo a double displacement reaction to form silver cyanide (AgCN) and potassium nitrate (KNO3). Hence the unbalanced chemical equation is:\nKCN(aq) + AgNO3(aq) → AgCN(s) + KNO3(aq).\nThe equation is already balanced.
02

(b) Combustion of Si3H8 in oxygen

The reactants for this reaction are the silicon hydride (Si3H8) and oxygen (O2). In a typical combustion reaction with an excess of oxygen, the products are expected to be silicon dioxide (SiO2) and water (H2O). The unbalanced reaction is: \nSi3H8(g) + O2(g) → SiO2(s) + H2O(g).\nBalancing the equation gives: \n3Si3H8(g) + 16O2(g) → 9SiO2(s) + 12H2O(g).
03

(c) Reaction of dinitrogen with calcium carbide

For this reaction, dinitrogen (N2) reacts with calcium carbide (CaC2) to yield calcium cyanamide (CaNCN). The nitrogen atom in N2 is triple-bonded, highly un-reactive and requires a significant amount of energy to break. It is usually performed in presence of a catalyst under high heat and pressure but for simplicity's sake, we'll ignore these details. The unbalanced reaction is: \nN2(g) + CaC2(s) → CaNCN(s).\nBalancing this equation gives: \nN2(g) + 3CaC2(s) → 3CaNCN(s) + C(s). Note the addition of Carbon (C) as a product to balance the carbons on both sides.

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Most popular questions from this chapter

The molecule tetraborane has the formula \(\mathrm{B}_{4} \mathrm{H}_{10}\) (a) Show that this is an electron-deficient molecule. (b) How many bridge bonds must occur in the molecule? (c) Show that butane, \(\mathrm{C}_{4} \mathrm{H}_{10},\) is not electron deficient.

Would you expect the reaction of \(\mathrm{Pb}(\mathrm{s})\) and \(\mathrm{Cl}_{2}(\mathrm{g})\) to yield \(\mathrm{PbCl}_{2}\) or \(\mathrm{PbCl}_{4} ?\)

Write the simplest chemical equation to represent the reaction of (a) \(\mathrm{K}_{2} \mathrm{CO}_{3}\left(\text { aq) and } \mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq})\right.\); (b) \(\operatorname{Mg}\left(\mathrm{HCO}_{3}\right)_{2}(\) aq) on heating; (c) tin(II) oxide when heated with carbon; (d) \(\mathrm{CaF}_{2}(\mathrm{s})\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{concd}\) aq); (e) \(\mathrm{NaHCO}_{3}(\mathrm{s})\) and \(\mathrm{HCl}(\mathrm{aq}) ;\) (f) \(\mathrm{PbO}_{2}(\mathrm{s})\) and HBr(aq); and (g) the reduction of \(\mathrm{SiF}_{4}\) to pure \(\mathrm{Si},\) by using Na as the reducing agent.

An analysis of a Solvay-process plant shows that for every \(1.00 \mathrm{kg}\) of \(\mathrm{NaCl}\) consumed, \(1.03 \mathrm{kg}\) of \(\mathrm{NaHCO}_{3}\) are obtained. The quantity of \(\mathrm{NH}_{3}\) consumed in the overall process is \(1.5 \mathrm{kg}.\) (a) What is the percent efficiency of this process for converting NaCl to \(\mathrm{NaHCO}_{3} ?\) (b) Why is so little \(\mathrm{NH}_{3}\) required?

Which has the (a) higher melting point, MgO or BaO; (b) greater solubility in water, \(\mathrm{MgF}_{2}\) or \(\mathrm{MgCl}_{2}\) ? Explain.
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