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Chapter 7: Problem 69

Write a balanced equation for the reaction that occurs in each of the following cases: (a) Potassium metal is exposed to an atmosphere of chlorine gas. (b) Strontium oxide is added to water. (c) A fresh surface of lithium metal is exposed to oxygen gas. (d) Sodium metal reacts with molten sulfur.

Short Answer

Expert verified
The balanced equations for the given reactions are: (a) \(2K + Cl_2 \rightarrow 2KCl\) (b) \(SrO + H_2O \rightarrow Sr(OH)_2\) (c) \(4Li + O_2 \rightarrow 2Li_2O\) (d) \(2Na + S \rightarrow Na_2S\)

Step by step solution

01

(a) Potassium metal reacts with chlorine gas

To write the balanced equation for the reaction between potassium metal (K) and chlorine gas (Cl₂), first identify the products. Potassium belongs to group I, hence it has +1 charge, and chlorine belongs to group VII, hence it has -1 charge. So, the product will be potassium chloride (KCl). Now, balance the equation: 2K + Cl₂ → 2KCl The balanced equation is: \(2K + Cl_2 \rightarrow 2KCl\)
02

(b) Strontium oxide reacts with water

To write the balanced equation for the reaction between strontium oxide (SrO) and water (H₂O), first identify the products. SrO has a +2 charge on Sr and -2 charge on O; water has +1 charge on H and -2 charge on O. The product will be strontium hydroxide (Sr(OH)₂), as Sr has a +2 charge and OH⁻ has -1 charge. Now, balance the equation: SrO + H₂O → Sr(OH)₂ The balanced equation is: \(SrO + H_2O \rightarrow Sr(OH)_2\)
03

(c) Lithium metal reacts with oxygen gas

To write the balanced equation for the reaction between lithium metal (Li) and oxygen gas (O₂), first identify the products. Lithium belongs to group I, hence it has +1 charge, and oxygen belongs to group VI, hence it has -2 charge. So, the product will be lithium oxide (Li₂O). Now, balance the equation: 4Li + O₂ → 2Li₂O The balanced equation is: \(4Li + O_2 \rightarrow 2Li_2O\)
04

(d) Sodium metal reacts with molten sulfur

To write the balanced equation for the reaction between sodium metal (Na) and molten sulfur (S), first identify the products. Sodium belongs to group I, hence it has +1 charge, and sulfur belongs to group VI, hence it has -2 charge. So, the product will be sodium sulfide (Na₂S). Now, balance the equation: 2Na + S → Na₂S The balanced equation is: \(2Na + S \rightarrow Na_2S\)

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Most popular questions from this chapter

The following observations are made about two hypothetical elements \(\mathrm{A}\) and \(\mathrm{B}:\) The \(\mathrm{A}-\mathrm{A}\) and \(\mathrm{B}-\mathrm{B}\) bond lengths in the elemental forms of \(\mathrm{A}\) and \(\mathrm{B}\) are 236 and \(194 \mathrm{pm}\), respectively. A and \(B\) react to form the binary compound \(A B_{2}\), which has a linear structure (that is \(\left.\angle \mathrm{B}-\mathrm{A}-\mathrm{B}=180^{\circ}\right) .\) Based on these statements, predict the separation between the two B nuclei in a molecule of \(\mathrm{AB}_{2}\).

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