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Chapter 20: Problem 20

Write balanced equations describing the reaction of \(\mathrm{Sr}\) with each of the following: \(\mathrm{O}_{2}, \mathrm{~S}, \mathrm{Cl}_{2}, \mathrm{P}_{4}, \mathrm{H}_{2}, \mathrm{H}_{2} \mathrm{O}\), and \(\mathrm{HCl}\).

Short Answer

Expert verified
The balanced equations for the reactions of strontium (Sr) with various substances are as follows: 1. With oxygen (O鈧�): \( Sr + O_2 \rightarrow SrO \) 2. With sulfur (S): \( Sr + S \rightarrow SrS \) 3. With chlorine (Cl鈧�): \( Sr + Cl_2 \rightarrow SrCl_2 \) 4. With phosphorus (P鈧�): \( 3Sr + P_4 \rightarrow 2Sr_3P_2 \) 5. With hydrogen (H鈧�): \( Sr + H_2 \rightarrow SrH_2 \) 6. With water (H鈧侽): \( Sr + 2H_2O \rightarrow Sr(OH)_2 + H_2 \) 7. With hydrochloric acid (HCl): \( Sr + 2HCl \rightarrow SrCl_2 + H_2 \)

Step by step solution

01

Reaction of Sr with O鈧�

Strontium is a group 2 element, so it has a valency of +2. Oxygen is a group 16 element, with a valency of -2. Therefore, when they form a compound, the ratio between the two elements will be 1:1. The balanced equation is: \[ Sr + O_2 \rightarrow SrO \]
02

Reaction of Sr with S

Sulfur, like oxygen, is a group 16 element with a valency of -2. Thus, when Sr and S form a compound, their ratio will also be 1:1. The balanced equation is: \[ Sr + S \rightarrow SrS \]
03

Reaction of Sr with Cl鈧�

Chlorine is a group 17 element, with a valency of -1. Strontium has a valency of +2, so when they form a compound the ratio of the elements will be 1 Sr to 2 Cl. The balanced equation is: \[ Sr + Cl_2 \rightarrow SrCl_2 \]
04

Reaction of Sr with P鈧�

Phosphorus is a group 15 element with a valency of -3. To balance the valency of strontium, which is +2, we need to use 3 Sr and 2 P to form the compound Sr鈧働鈧�. The balanced equation is: \[ 3Sr + P_4 \rightarrow 2Sr_3P_2 \]
05

Reaction of Sr with H鈧�

Hydrogen is a group 1 element with a valency of +1. When it forms a compound with strontium it takes the form of SrH鈧� because it has to balance the +2 valency of Sr. The balanced equation is: \[ Sr + H_2 \rightarrow SrH_2 \]
06

Reaction of Sr with H鈧侽

When Sr reacts with water, it forms Sr(OH)鈧� and hydrogen gas (H鈧�) is released. The balanced equation is: \[ Sr + 2H_2O \rightarrow Sr(OH)_2 + H_2 \]
07

Reaction of Sr with HCl

Hydrochloric acid (HCl) is an acid with a chloride (Cl鈦�) ion and H鈦� ion. When Sr reacts with HCl, it forms SrCl鈧� and hydrogen gas (H鈧�) is released. The balanced equation is: \[ Sr + 2HCl \rightarrow SrCl_2 + H_2 \]

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