What mass of oxygen will react with 10.5g of ethane according to the following reaction?
2 C2H6 + 7 O2 = 4 CO2 + 6 H2O

What mass of carbon will react with 15.7g of iron(III)oxide according to the following reaction?
2Fe2O3 + 3C = 4Fe + 3CO2

What mass of iron(III) oxide is required to produce 16.4g of iron?
2Fe2O3 + 3C → 4Fe + 3CO2

What mass of oxygen will react with 17.5 g of aluminum according to the following reaction?
4 Al + 3 O2 → 2 Al2O3

What mass of sodium nitride is required to produce 15.1g of nitrogen?
2Na3N → N2 + 6Na

Now compare them and find the moles of the substance whose mass you need using molar ratio. Use the formula written above to find the theoretical Mass required for reaction.

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pstnonsonjoku pstnonsonjoku · Answer 2 · 2021-11-06T16:10:58+01:00

The amount of reactants or products can be obtained from the stoichiometry of the reaction.

The reaction is shown by the equation;

2C2H6 + 7O2 ---------> 4CO2 + 6H2O

Number of moles of ethane = 10.5g/30g/mol = 0.35 moles

2 moles ethane reacts with 7 moles of oxygen

0.35 moles of ethane reacts with 0.35 moles × 7 moles/2 moles

= 1.225 moles of oxygen

Mass of oxygen = 1.225 moles of oxygen × 32 g/mol

= 39.2 g

The equation of the reaction is;

2Fe2O3 + 3C-------> 4Fe + 3CO2

Number of moles of iron(III)oxide = 15.7g /160g/mol = 0.098 moles

2 moles of iron(III)oxide reacts with 3 moles of carbon

0.098 moles of iron(III)oxide reacts with 0.098 moles × 3 moles/2 moles

= 0.147 moles

Mass of C= 0.147 moles × 12 g/mol = 1.764 g

The equation of the reaction is;

2Fe2O3 + 3C → 4Fe + 3CO2

Number of moles of iron = 16.4g/56 g/mol = 0.29 moles

2 moles of iron(III) oxide produces 4 moles of iron

x moles of iron(III) oxide produces 0.29 moles of iron

x = 2 moles × 0.29 moles/4 moles

x = 0.145 moles

Mass of iron(III) oxide = 0.145 moles × 160g/mol

= 23.2 g

The equation of the reaction is;

4 Al + 3 O2 → 2 Al2O3

Number of moles of Al = 17.5 g /27 g/mol = 0.65 moles

4 moles of Al reacts with 3 moles of oxygen

0.65 moles reacts with 0.65 moles × 3 moles/4 moles

= 0.49 moles

Mass of oxygen = 0.49 moles × 32 g/mol = 15.68 g

Equation of the reaction is;

2Na3N → N2 + 6Na

Number of moles of nitrogen = 15.1g/28 g/mol = 0.54 moles

2 moles of sodium nitride produces 1 mole of N2

x moles of sodium nitride produces 0.54 moles of N2

x = 2 moles × 0.54 moles /1 mole

x = 1.08 moles

Mass of sodium nitride = 1.08 moles × 83 g/mol

= 89.64 g

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