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Chapter 4: Q87E (page 224)

What volume of 0.600 M HCl is required to react completely with 2.50 g of sodium hydrogen carbonate?

\({\rm{NaHC}}{{\rm{O}}_{3(aq)}}{\rm{ + HC}}{{\rm{l}}_{(aq)}}{\rm{ }} \to {\rm{ NaC}}{{\rm{l}}_{(aq)}}{\rm{ + C}}{{\rm{O}}_{2(aq)}}{\rm{ + }}{{\rm{H}}_2}{{\rm{O}}_{({\rm{l}})}}\)

Short Answer

Expert verified

50 ml of hydrochloric acid will be required to react with 0.2.50 g sodium hydrogen carbonate.

Step by step solution

01

Number of moles of NaHCO3

The number of moles is the ratio of the given mass and molar mass of the compound. As the given mass of \({\rm{NaHC}}{{\rm{O}}_3}\)is 2.50 g and molecular mass of \({\rm{NaHC}}{{\rm{O}}_3}\)is \({\rm{23 + 1 + 12 + 16 }} \times {\rm{ 3}} = {\rm{ 84 g/mol}}\)

Therefore, the number of moles = \(\frac{{2.50}}{{84}}{\rm{ = 0}}{\rm{.03 moles}}\)

Now, as 1 moles of hydrochloric acid is required to react with one mole of \({\rm{NaHC}}{{\rm{O}}_3}\), thus the number of moles of nitric acid will be \({\rm{0}}{\rm{.03 }} \times {\rm{ 1 = 0}}{\rm{.03 moles}}\)

02

Volume of hydrochloric acid

The volume of hydrochloric acid, according to the mole concept can be calculated as:

\({\rm{Concentration = }}\frac{{{\rm{Moles}}}}{{{\rm{Volume}}}}\)

It is given that the concentration of hydrochloric acid is 0.600 M and the number of moles are 0.03 moles, therefore volume of hydrochloric acid will be

\({\rm{Volume = }}\frac{{0.03}}{{0.600}}{\rm{ = 50 mL}}\)

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Most popular questions from this chapter

Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.

(a)\({H_3}P{O_4}\)

(b)\(Al{\left( {OH} \right)_3}\)

(c)\(Se{O_2}\)

(d)\(KN{O_2}\)

(e)\(I{n_2}{S_3}\)

(f)\({P_4}{O_6}\).

How many milliliters of a 0.1500-M solution of KOH will be required to titrate 40.00 mL of a 0.0656-M solution of H3PO4?

\({{\rm{H}}_3}{\rm{P}}{{\rm{O}}_{4({\rm{aq}})}}{\rm{ + 2KO}}{{\rm{H}}_{({\rm{aq}})}}{\rm{ }} \to {\rm{ }}{{\rm{K}}_2}{\rm{HP}}{{\rm{O}}_{4({\rm{aq}})}}{\rm{ + 2}}{{\rm{H}}_2}{{\rm{O}}_{({\rm{l}})}}\)

Balance the following equations:

\(\begin{array}{l}\;(a)\;PC{l_5}\left( s \right) + {H_2}O\left( l \right) \to POC{l_3}\left( l \right) + HCl\left( {aq} \right)\\\left( b \right)\,Cu\left( s \right) + HN{O_3}\left( {aq} \right) \to Cu{\left( {N{O_3}} \right)_2}\left( {aq} \right) + {H_2}O\left( l \right) + NO\left( g \right)\\\left( c \right){H_2}\left( g \right) + {I_2}\left( g \right) \to HI\left( s \right)\\\left( d \right)\,Fe\left( s \right) + {O_2}\left( g \right) \to F{e_2}{O_3}\left( s \right)\\\left( e \right)\,Na\left( s \right) + {H_2}O\left( l \right) \to NaOH\left( {aq} \right) + {H_2}\left( g \right)\\\left( f \right)\,{\left( {N{H_4}} \right)_2}C{r_2}{O_7}\left( s \right) \to C{r_2}{O_3}\left( s \right) + {N_2}\left( g \right) + {H_2}O\left( g \right)\\\left( g \right){P_4}\left( s \right) + C{l_2}\left( g \right) \to PC{l_3}\left( l \right)\\\left( h \right)PtC{l_4}\left( s \right) \to Pt\left( s \right) + C{l_2}\left( g \right)\end{array}\)

A 0.025-g sample of a compound composed of boron and hydrogen, with a molecular mass of ~28 amu, burns spontaneously when exposed to air, producing 0.063 g of B2O3. What are the empirical and molecular formulas of the compound?

What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503MAgNO3 requires 20.22 mL of the AgNO3 solution to reach the end point?

AgNO3(aq) + NaCl(aq)鉄禔驳颁濒(s) + NaNO3(aq)
See all solutions

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