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Chapter 4: Problem 2

Balance the following equations by inspection. (a) \(\mathrm{P}_{2} \mathrm{H}_{4} \longrightarrow \mathrm{PH}_{3}+\mathrm{P}_{4}\) (b) \(\mathrm{P}_{4}+\mathrm{Cl}_{2} \longrightarrow \mathrm{PCl}_{3}\) (c) \(\mathrm{FeCl}_{3}+\mathrm{H}_{2} \mathrm{S} \longrightarrow \mathrm{Fe}_{2} \mathrm{S}_{3}+\mathrm{HCl}\) (d) \(\mathrm{Mg}_{3} \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Mg}(\mathrm{OH})_{2}+\mathrm{NH}_{3}\)

Short Answer

Expert verified
For equation (a): \(P_{2}H_{4} \longrightarrow 2PH_{3}+P_{4}\). For equation (b): \(P_{4}+6Cl_{2} \longrightarrow 4PCl_{3}\). For equation (c): \(2FeCl_{3}+3H_{2}S \longrightarrow Fe_{2}S_{3}+6HCl\). For equation (d): \(Mg_{3}N_{2}+6H_{2}O \longrightarrow 3Mg(OH)_{2}+2NH_{3}\).

Step by step solution

01

Balance equation (a)

To balance this equation, start with the phosphorus (P) atoms. On the left-hand side, there are 2 P atoms while on the right-hand side there's 1 in PH3 and 4 in P4, making a total of 5. To resolve this, change the coefficient in front of PH3 to 4. Now, there are 2*4 = 8 H atoms in P2H4 on the left and 3*4 = 12 H in PH3 on the right. To fix this, add a coefficent of 2 in front of PH3, giving the balance equation as: \[P_{2}H_{4} \longrightarrow 2PH_{3}+P_{4}\]
02

Balance equation (b)

Start with the phosphorus atoms. Both sides have 4 P atoms, which are already balanced. Now, see the chlorine (Cl) atoms. On the left, there are 2 Cl whereas on the right there are 3*4 = 12 Cl. Balance this by placing a 6 in front of Cl2 on the left: \[P_{4}+6Cl_{2} \longrightarrow 4PCl_{3}\]
03

Balance equation (c)

Start with Iron (Fe) atoms. Place a coefficient of 2 in front of FeCl3 on the left to balance with the 2 Fe on the right. Now, see the sulfur (S) atoms. We have 1 S on the left and 3 on the right. Balance this by placing a 3 in front of H2S on the left side. Finally, balance the Hydrogen (H) and Chlorine (Cl) atoms by placing a 6 in front of HCl on the right side: \[2FeCl_{3}+3H_{2}S \longrightarrow Fe_{2}S_{3}+6HCl\]
04

Balance equation (d)

Begin with magnesium (Mg) atoms. To match the 3 Mg atoms on the left with those on the right, place a coefficient of 3 in front of Mg(OH)2 on the right side. There are 2 nitrogen (N) atoms on the left and 1 on the right. Balance them by placing a 2 in front of NH3 on the right side. The Hydrogen atoms are already balanced! Lastly, balance the oxygen atoms by place a 6 in front of H2O on the left: \[Mg_{3}N_{2}+6H_{2}O \longrightarrow 3Mg(OH)_{2}+2NH_{3}\]

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Most popular questions from this chapter

The reaction of calcium hydride with water can be used to prepare small quantities of hydrogen gas, as is done to fill weather-observation balloons. \(\mathrm{CaH}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow\) $$ \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g}) \text { (not balanced) } $$ (a) How many grams of \(\mathrm{H}_{2}(\mathrm{g})\) result from the reaction of \(127 \mathrm{g} \mathrm{CaH}_{2}\) with an excess of water? (b) How many grams of water are consumed in the reaction of \(56.2 \mathrm{g} \mathrm{CaH}_{2} ?\) (c) What mass of \(\mathrm{CaH}_{2}(\mathrm{s})\) must react with an excess of water to produce \(8.12 \times 10^{24}\) molecules of \(\mathrm{H}_{2} ?\)

Consider the reaction below. \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{CaCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (a) How many grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) are required to react completely with \(415 \mathrm{mL}\) of \(0.477 \mathrm{M} \mathrm{HCl} ?\) (b) How many kilograms of \(\mathrm{Ca}(\mathrm{OH})_{2}\) are required to react with 324 L of a HCl solution that is 24.28\% HCl by mass, and has a density of \(1.12 \mathrm{g} / \mathrm{mL} ?\)

Chromium(II) sulfate, \(\mathrm{CrSO}_{4^{\prime}}\) is a reagent that has been used in certain applications to help reduce carbon-carbon double bonds \((\mathrm{C}=\mathrm{C})\) in molecules to single bonds ( \(\mathrm{C}-\mathrm{C}\) ). The reagent can be prepared via the following reaction. $$\begin{array}{c} 4 \mathrm{Zn}(\mathrm{s})+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{aq})+7 \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \longrightarrow 4 \mathrm{ZnSO}_{4}(\mathrm{aq})+2 \mathrm{CrSO}_{4}(\mathrm{aq})+\mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})+7 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$

A \(5.00 \mathrm{mL}\) sample of an aqueous solution of \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\) requires \(49.1 \mathrm{mL}\) of \(0.217 \mathrm{M} \mathrm{NaOH}\) to convert all of the \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\) to \(\mathrm{Na}_{3} \mathrm{PO}_{4} .\) The other product of the reaction is water. Calculate the molarity of the \(\mathrm{Na}_{2} \mathrm{H}_{3} \mathrm{PO}_{4}\) solution.

The manufacture of ethyl alcohol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) yields diethyl ether, \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}\) as a by-product. The complete combustion of a \(1.005 \mathrm{g}\) sample of the product of this process yields \(1.963 \mathrm{g} \mathrm{CO}_{2} .\) What must be the mass percents of \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right),\) and of \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}\) in this sample?
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