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Chapter 22: Problem 23

Complete and balance the following equations: (a) \(\mathrm{NaH}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (b) \(\mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow\) (d) \(\mathrm{Na}(l)+\mathrm{H}_{2}(g) \longrightarrow\) (e) \(\mathrm{PbO}(s)+\mathrm{H}_{2}(g) \longrightarrow\)

Short Answer

Expert verified
The short answer for the given chemical reactions is as follows: (a) \( \mathrm{NaH}(s) + \mathrm{H}_2\mathrm{O}(l) \longrightarrow \mathrm{NaOH}(s) + \mathrm{H}_2(g) \) (b) \( \mathrm{Fe}(s) + \mathrm{H}_{2}\mathrm{SO}_4(a q) \longrightarrow \mathrm{FeSO}_4(s) + \mathrm{H}_2(g) \) (c) \( \mathrm{H}_2(g) + \mathrm{Br}_2(g) \longrightarrow 2\mathrm{HBr}(g) \) (d) \( \mathrm{Na}(l) + \mathrm{H}_{2}(g) \longrightarrow \mathrm{NaH}(s) + \mathrm{H}_2(g) \) (e) \( \mathrm{PbO}(s) + \mathrm{H}_{2}(g) \longrightarrow \mathrm{Pb}(s) + \mathrm{H}_2\mathrm{O}(l) \)

Step by step solution

01

Identify the Reaction Type and Products

This reaction is an acid-base reaction where sodium hydride (NaH) acts as a strong base and water (H2O) acts as an acid. The reaction results in the formation of sodium hydroxide (NaOH) and hydrogen gas (H2).
02

Write the Unbalanced Equation

Now, let's write the products of the reaction: \[\mathrm{NaH}(s) + \mathrm{H}_2\mathrm{O}(l) \longrightarrow \mathrm{NaOH}(s) + \mathrm{H}_2(g)\]
03

Balance the Equation

Both sides of the equation already have equal numbers of atoms for each element (1 Na, 2 H, and 1 O). Therefore, the equation is balanced as written. #a) Balanced Equation#: \( \mathrm{NaH}(s) + \mathrm{H}_2\mathrm{O}(l) \longrightarrow \mathrm{NaOH}(s) + \mathrm{H}_2(g) \) #b)# Complete and Balance the Equation: \(\mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\)
04

Identify the Reaction Type and Products

This is a single replacement reaction where the metal (Fe) reacts with an acid (H2SO4). The reaction results in the formation of iron(II) sulfate (FeSO4) and hydrogen gas (H2).
05

Write the Unbalanced Equation

Now, let's write the products of the reaction: \[\mathrm{Fe}(s) + \mathrm{H}_2\mathrm{SO}_4(aq) \longrightarrow \mathrm{FeSO}_4(s) + \mathrm{H}_2(g)\]
06

Balance the Equation

Since there is only 1 atom of each element on both sides of the equation, it is already balanced. #b) Balanced Equation#: \( \mathrm{Fe}(s) + \mathrm{H}_{2}\mathrm{SO}_4(a q) \longrightarrow \mathrm{FeSO}_4(s) + \mathrm{H}_2(g) \) #c)# Complete and Balance the Equation: \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow\)
07

Identify the Reaction Type and Products

This is a synthesis reaction where hydrogen gas (H2) and bromine gas (Br2) combine to form hydrogen bromide (HBr).
08

Write the Unbalanced Equation

Now, let's write the products of the reaction: \[\mathrm{H}_2(g) + \mathrm{Br}_2(g) \longrightarrow \mathrm{HBr}(g)\]
09

Balance the Equation

To balance the equation, we need 2 HBr molecules on the product side: \[\mathrm{H}_2(g) + \mathrm{Br}_2(g) \longrightarrow 2\mathrm{HBr}(g)\] #c) Balanced Equation#: \( \mathrm{H}_2(g) + \mathrm{Br}_2(g) \longrightarrow 2\mathrm{HBr}(g) \) #d)# Complete and Balance the Equation: \(\mathrm{Na}(l)+\mathrm{H}_{2}(g) \longrightarrow\)
10

Identify the Reaction Type and Products

This is a single replacement reaction where the metal sodium (Na) reacts with hydrogen gas (H2), leading to the formation of sodium hydride (NaH) and hydrogen gas (H2).
11

Write the Unbalanced Equation

Now, let's write the products of the reaction: \[\mathrm{Na}(l) + \mathrm{H}_2(g) \longrightarrow \mathrm{NaH}(s) + \mathrm{H}_2(g)\]
12

Balance the Equation

Both sides of the equation have equal numbers of atoms for each element, making the equation balanced as written. #d) Balanced Equation#: \( \mathrm{Na}(l) + \mathrm{H}_{2}(g) \longrightarrow \mathrm{NaH}(s) + \mathrm{H}_2(g) \) #e)# Complete and Balance the Equation: \(\mathrm{PbO}(s)+\mathrm{H}_{2}(g) \longrightarrow\)
13

Identify the Reaction Type and Products

This is a single replacement reaction between a metal oxide (PbO) and hydrogen gas (H2). The reaction forms lead (Pb) and water (H2O).
14

Write the Unbalanced Equation

Now, let's write the products of the reaction: \[\mathrm{PbO}(s) + \mathrm{H}_2(g) \longrightarrow \mathrm{Pb}(s) + \mathrm{H}_2\mathrm{O}(l)\]
15

Balance the Equation

Both sides of the equation have equal numbers of atoms for each element, so the equation is balanced as written. #e) Balanced Equation#: \( \mathrm{PbO}(s) + \mathrm{H}_{2}(g) \longrightarrow \mathrm{Pb}(s) + \mathrm{H}_2\mathrm{O}(l) \)

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Most popular questions from this chapter

Write the chemical formula for each of the following compounds, and indicate the oxidation state of the group 16 elements in each: (a) selenious acid, (b) sulphur trioxide, \((\mathbf{c})\) selenium dichloride, \((\mathbf{d})\) aluminium selenide, (e) iron(II) sulfate, (f) tellurium trioxide.

Write the chemical formula for each of the following, and indicate the oxidation state of the halogen or noble-gas atom in each: (a) krypton tetrafluoride, (b) hexafluoroantimonate ion, \((\mathbf{c})\) sodium hypoiodite, \((\mathbf{d})\) perbromic acid, (e) aluminum perchlorate, (f) iron(II) iodite.

Borazine, \((\mathrm{BH})_{3}(\mathrm{NH})_{3},\) is an analog of \(\mathrm{C}_{6} \mathrm{H}_{6},\) benzene. It can be prepared from the reaction of diborane with ammonia, with hydrogen as another product; or from lithium borohydride and ammonium chloride, with lithium chloride and hydrogen as the other products. (a) Write balanced chemical equations for the production of borazine using both synthetic methods. (b) Draw the Lewis dot structure of borazine. (c) How many grams of borazine can be prepared from $2.00 \mathrm{~L}$ of ammonia at STP, assuming diborane is in excess?

Identify the true statements concerning the atoms and ions of the group 16 elements. [Sections 22.5 and 22.6\(]\) (a) The ionic radii are larger than the atomic radii because the ions have more electrons than their corresponding atoms. (b) Atomic radii increase going down the group because of increasing nuclear charge. (c) The ionic radii increase going down the group because of the increase in the principal quantum number of outermost electrons. (d) Of these ions, \(\mathrm{Se}^{2-}\) is the strongest base in water because it is largest.

What is the anhydride for each of the following acids: (a) $\mathrm{H}_{2} \mathrm{SO}_{4},(\mathbf{b}) \mathrm{HClO}_{3},(\mathbf{c}) \mathrm{HNO}_{2},\( (d) \)\mathrm{H}_{2} \mathrm{CO}_{3},(\mathbf{e}) \mathrm{H}_{3} \mathrm{PO}_{4} ?$
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