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Chapter 4: Problem 50

Excess \(\mathrm{NaHCO}_{3}\) is added to \(525 \mathrm{mL}\) of \(0.220 \mathrm{M}\) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} .\) These substances react as follows: \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{NaHCO}_{3}(\mathrm{s}) \longrightarrow\) $$ \mathrm{CuCO}_{3}(\mathrm{s})+2 \mathrm{NaNO}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) $$ (a) How many grams of the \(\mathrm{NaHCO}_{3}(\mathrm{s})\) will be consumed? (b) How many grams of \(\mathrm{CuCO}_{3}(\mathrm{s})\) will be produced?

Short Answer

Expert verified
(a) 19.41 g of NaHCO3 will be consumed. (b) 14.26 g of CuCO3 will be produced.

Step by step solution

01

Determine moles of copper nitrate [Cu(NO3)2]

We know the volume (0.525 L) and molarity (0.220 M) of the copper nitrate solution. Apply the equation ‘n=c×V’ where 'n' is the number of moles, 'c' is the molarity and 'V' is the volume in liter. Calculate as 'n = 0.220 M × 0.525 L = 0.1155 mol'. So, we have 0.1155 mol of copper nitrate.
02

Determine moles of sodium bicarbonate (NaHCO3) that will be consumed

From the balanced chemical equation, we see that 1 mol of Cu(NO3)2 reacts with 2 mol of NaHCO3. Therefore, the amount of NaHCO3 consumed will be twice the amount of Cu(NO3)2. Calculate as 'n(NaHCO3) = 2 * n(Cu(NO3)2) = 2 * 0.1155 mol = 0.231 mol'
03

Compute the mass of NaHCO3 consumed

Convert the moles of NaHCO3 to grams using its molar mass (84 g/mol). Use the equation ‘m = n × M’ where 'm' is the mass, 'n' is the number of moles and 'M' is the molar mass. Calculate as 'm(NaHCO3) = 0.231 mol * 84 g/mol = 19.41 g'
04

Determine moles of copper carbonate (CuCO3) that will be produced

From the balanced chemical equation, 1 mol of Cu(NO3)2 produces 1 mol CuCO3. So the amount of CuCO3 produced will be the same as the amount of Cu(NO3)2. Therefore, n(CuCO3) = n(Cu(NO3)2) = 0.1155 mol.
05

Compute the mass of CuCO3 produced

Convert the moles of CuCO3 to grams using its molar mass (123.5 g/mol). Calculate as 'm(CuCO3) = n(CuCO3) * M(CuCO3) = 0.1155 mol * 123.5 g/mol = 14.26 g'

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Most popular questions from this chapter

Iron ore is impure \(\mathrm{Fe}_{2} \mathrm{O}_{3} .\) When \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is heated with an excess of carbon (coke), metallic iron and carbon monoxide gas are produced. From a sample of ore weighing \(938 \mathrm{kg}, 523 \mathrm{kg}\) of pure iron is obtained. What is the mass percent \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) by mass, in the ore sample, assuming that none of the impurities contain Fe?

High-purity silicon is obtained using a three-step process. The first step involves heating solid silicon dioxide, \(\mathrm{SiO}_{2^{\prime}}\) with solid carbon to give solid silicon and carbon monoxide gas. In the second step, solid silicon is converted into liquid silicon tetrachloride, \(\mathrm{SiCl}_{4}\) by treating it with chlorine gas. In the last step, \(\mathrm{SiCl}_{4}\) is treated with hydrogen gas to give ultrapure solid silicon and hydrogen chloride gas. (a) Write balanced chemical equations for the steps involved in this three- step process. (b) Calculate the masses of carbon, chlorine, and hydrogen required per kilogram of silicon.

Cryolite, \(\mathrm{Na}_{3} \mathrm{AlF}_{6^{\prime}}\) is an important industrial reagent. It is made by the reaction below. $$\begin{array}{r} \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+6 \mathrm{NaOH}(\mathrm{aq})+12 \mathrm{HF}(\mathrm{g}) \longrightarrow 2 \mathrm{Na}_{3} \mathrm{AlF}_{6}(\mathrm{s})+9 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$ In an experiment, \(7.81 \mathrm{g} \mathrm{Al}_{2} \mathrm{O}_{3}\) and excess \(\mathrm{HF}(\mathrm{g})\) were dissolved in 3.50 L of 0.141 M NaOH. If 28.2 g \(\mathrm{Na}_{3} \mathrm{AlF}_{6}\) was obtained, then what is the percent yield for this experiment?

Exactly \(1.00 \mathrm{mL}\) of an aqueous solution of \(\mathrm{HNO}_{3}\) is diluted to \(100.0 \mathrm{mL}\). It takes \(29.78 \mathrm{mL}\) of \(0.0142 \mathrm{M}\) \(\mathrm{Ca}(\mathrm{OH})_{2}\) to convert all of the \(\mathrm{HNO}_{3}\) to \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) The other product of the reaction is water. Calculate the molarity of the undiluted HNO \(_{3}\) solution.

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 \(1.12 \mathrm{g} / \mathrm{mL} ?\)
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