A sample of 15 . 2 g of tin(IV) oxide is mixed with 2 . 41 g of carbon and heated. A reaction occurs. SnO 2 + 2 C → Sn + 2 CO Show by calculation that the reagent in excess is tin(IV) oxide. ( A r values: Sn = 118 . 7 , C = 12 . 0 , O = 16 . 0 ).

In a chemical reaction, reactants that are not used up when the reaction is finished are called excess reagents. To find the excess reagent we have to calculate the moles number of each reactant for the given reaction by using stoichiometric reaction, SnO 2 + 2 C → Sn + 2 CO From the above reaction it is confirmed that 1 mole of tin IV oxide reacts with 2 moles of carbon to give 1 mole of tin and 2 moles of carbon monoxide. In the problem they have given, Weight of tin IV oxide = 15 . 2 g Molar mass of tin IV oxide = 150 . 7 g / mol Then, moles number of tin IV oxide = 15 . 2 150 . 7 = 0 . 1009 mol Now we will calculate moles of carbon required for complete consumption of 0 . 1009 mol tin IV oxide by using conversion factor, = 0 . 1009 mol × 2 mol of C 1 mol of SnO 2 = 0 . 1009 × 2 1 = 0 . 2018 mol of carbon needed. Weight of carbon = 2 . 41 g Then, moles number of carbon = 2 . 41 12 . 0 = 0 . 2008 mol of carbon reacted So, from the above values for 0 . 2018 mol of carbon only 0 . 2008 mol carbon is added, this indicates SnO 2 is excess reagent.

A sample of 15 . 2 g of tin(IV) oxide is mixed with 2 . 41 g of carbon and heated. A reaction occurs. SnO 2 + 2 C → Sn + 2 CO Show by calculation that the reagent in excess is tin(IV) oxide. ( A r values: Sn = 118 . 7 , C = 12 . 0 , O = 16 . 0 ).

In a chemical reaction, reactants that are not used up when the reaction is finished are called excess reagents. To find the excess reagent we have to calculate the moles number of each reactant for the given reaction by using stoichiometric reaction, SnO 2 + 2 C → Sn + 2 CO From the above reaction it is confirmed that 1 mole of tin IV oxide reacts with 2 moles of carbon to give 1 mole of tin and 2 moles of carbon monoxide. In the problem they have given, Weight of tin IV oxide = 15 . 2 g Molar mass of tin IV oxide = 150 . 7 g / mol Then, moles number of tin IV oxide = 15 . 2 150 . 7 = 0 . 1009 mol Now we will calculate moles of carbon required for complete consumption of 0 . 1009 mol tin IV oxide by using conversion factor, = 0 . 1009 mol × 2 mol of C 1 mol of SnO 2 = 0 . 1009 × 2 1 = 0 . 2018 mol of carbon needed. Weight of carbon = 2 . 41 g Then, moles number of carbon = 2 . 41 12 . 0 = 0 . 2008 mol of carbon reacted So, from the above values for 0 . 2018 mol of carbon only 0 . 2008 mol carbon is added, this indicates SnO 2 is excess reagent.

A sample of zirconium is made by reacting 58 . 30 g of zirconium tetrachloride, ZrCl 4 , with excess magnesium. ZrCl 4 + 2 Mg → Zr + 2 MgCl 2 The mass of zirconium produced is 20 . 52 g . Calculate the percentage yield of zirconium. ( A r values: Zr = 91 . 2 , Cl = 35 . 5 , Mg = 24 . 3 ).

Given, actual mass of zirconium produced is 20 . 52 g Theoretical mass of zirconium is 91 . 2 g Percentage yield = Actual yield Theoretical yield × 100 % Percentage yield = 20 . 52 91 . 2 = 0 . 225 % .

Solid sodium carbonate reacts with aqueous hydrochloric acid to form aqueous sodium chloride, carbon dioxide and water. Na 2 CO 3 + 2 HCl → 2 NaCl + CO 2 + H 2 O Calculate the number of moles of hydrochloric acid required to react exactly with 4 . 15 g of sodium carbonate. ( A r values: C = 12 . 0 , Na = 23 . 0 , O = 16 . 0 , H = 1 . 0 , Cl = 35 . 5 ).

Solid sodium carbonate reacts with aqueous hydrochloric acid to form aqueous sodium chloride, carbon dioxide and water. Na 2 CO 3 + 2 HCl → 2 NaCl + CO 2 + H 2 O Molar mas of sodium carbonate, Na 2 CO 3 = 2 × 23 + 1 × 12 + 3 × 16 = 106 g / mol . Based on stoichiometric equation, 106 g of sodium carbonate reacts with 2 moles of hydrochloric acid. 4 . 15 g of sodium carbonate reacts with = 4 . 15 × 2 106 = 0 . 078 moles of hydrochloric acid.

EASY

AS and A Level

IMPORTANT

Earn 100

A sample of $15.2 g$ of tin(IV) oxide is mixed with $2.41 g$ of carbon and heated. A reaction occurs.

${SnO}_{2} + 2 C \to Sn + 2 CO$

Show by calculation that the reagent in excess is tin(IV) oxide. ( $A_{r}$ values: $Sn = 118.7, C = 12.0, O = 16.0$ ).

Important Questions on Atoms, Molecules and Stoichiometry

EASY

AS and A Level

IMPORTANT

Chemistry for Cambridge International AS & A Level Coursebook with Digital Access (2 Years)>Chapter 3 - Atoms, Molecules and Stoichiometry>EXAM-STYLE QUESTIONS>Q 2

A sample of zirconium is made by reacting $58.30 g$ of zirconium tetrachloride, ${ZrCl}_{4}$ , with excess magnesium.

${ZrCl}_{4} + 2 Mg \to Zr + 2 {MgCl}_{2}$

The mass of zirconium produced is $20.52 g$ . Calculate the percentage yield of zirconium. ( $A_{r}$ values: $Zr = 91.2, Cl = 35.5, Mg = 24.3$ ).

EASY

AS and A Level

IMPORTANT

Chemistry for Cambridge International AS & A Level Coursebook with Digital Access (2 Years)>Chapter 3 - Atoms, Molecules and Stoichiometry>EXAM-STYLE QUESTIONS>Q 3

Solid sodium carbonate reacts with aqueous hydrochloric acid to form aqueous sodium chloride, carbon dioxide and water.

${Na}_{2} {CO}_{3} + 2 HCl \to 2 NaCl + {CO}_{2} + H_{2} O$

Rewrite this equation to include state symbols.

EASY

AS and A Level

IMPORTANT

Chemistry for Cambridge International AS & A Level Coursebook with Digital Access (2 Years)>Chapter 3 - Atoms, Molecules and Stoichiometry>EXAM-STYLE QUESTIONS>Q 3

Solid sodium carbonate reacts with aqueous hydrochloric acid to form aqueous sodium chloride, carbon dioxide and water.

${Na}_{2} {CO}_{3} + 2 HCl \to 2 NaCl + {CO}_{2} + H_{2} O$

Calculate the number of moles of hydrochloric acid required to react exactly with $4.15 g$ of sodium carbonate.

( $A_{r}$ values: $C = 12.0, Na = 23.0, O = 16.0, H = 1.0, Cl = 35.5$ ).

EASY

AS and A Level

IMPORTANT

Chemistry for Cambridge International AS & A Level Coursebook with Digital Access (2 Years)>Chapter 3 - Atoms, Molecules and Stoichiometry>EXAM-STYLE QUESTIONS>Q 3

Define the term mole.

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AS and A Level

IMPORTANT

Chemistry for Cambridge International AS & A Level Coursebook with Digital Access (2 Years)>Chapter 3 - Atoms, Molecules and Stoichiometry>EXAM-STYLE QUESTIONS>Q 3

An aqueous solution of

25.0 {cm}^{3}

sodium carbonate of concentration

0.0200 mol {dm}^{- 3}

is titrated with hydrochloric acid. The volume of hydrochloric acid required to exactly react with the sodium carbonate is

12.50 {cm}^{3}

. Calculate the number of moles of sodium carbonate present in the solution of sodium carbonate.

EASY

AS and A Level

IMPORTANT

Chemistry for Cambridge International AS & A Level Coursebook with Digital Access (2 Years)>Chapter 3 - Atoms, Molecules and Stoichiometry>EXAM-STYLE QUESTIONS>Q 3

An aqueous solution of

25.0 {cm}^{3}

sodium carbonate of concentration

0.0200 mol {dm}^{- 3}

is titrated with hydrochloric acid. The volume of hydrochloric acid required to exactly react with the sodium carbonate is

12.50 {cm}^{3}

. Calculate the concentration of the hydrochloric acid.

EASY

AS and A Level

IMPORTANT

Chemistry for Cambridge International AS & A Level Coursebook with Digital Access (2 Years)>Chapter 3 - Atoms, Molecules and Stoichiometry>EXAM-STYLE QUESTIONS>Q 3

How many moles of carbon dioxide are produced when

0.2 mol

of sodium carbonate reacts with excess hydrochloric acid?

EASY

AS and A Level

IMPORTANT

Chemistry for Cambridge International AS & A Level Coursebook with Digital Access (2 Years)>Chapter 3 - Atoms, Molecules and Stoichiometry>EXAM-STYLE QUESTIONS>Q 3

Solid sodium carbonate reacts with aqueous hydrochloric acid to form aqueous sodium chloride, carbon dioxide and water.

${Na}_{2} {CO}_{3} + 2 HCl \to 2 NaCl + {CO}_{2} + H_{2} O$

Calculate the volume of this number of moles of carbon dioxide at r.t.p. ( $1 mol$ of gas occupies $24 {dm}^{3}$ at r.t.p.).

A sample of 15.2 g of tin(IV) oxide is mixed with 2.41 g of carbon and heated. A reaction occurs. SnO2+2C→Sn+2CO Show by calculation that the reagent in excess is tin(IV) oxide. (Ar values: Sn=118.7, C=12.0, O=16.0).

Important Questions on Atoms, Molecules and Stoichiometry

Learn and Prepare for any exam you want !

A sample of $15.2 g$ of tin(IV) oxide is mixed with $2.41 g$ of carbon and heated. A reaction occurs.

${SnO}_{2} + 2 C \to Sn + 2 CO$

Show by calculation that the reagent in excess is tin(IV) oxide. ( $A_{r}$ values: $Sn = 118.7, C = 12.0, O = 16.0$ ).