A 21.496 -g sample of magnesium is burned in air to form magnesium oxide and magnesium nitride.

step 1 We have given reaction is 2 of m .g plus o 2 into 2m .m .g. O 2m .O and 3 mol of m .g plus n2 in

A 21.496 -g sample of magnesium is burned in air to form...

Key Concepts

Stoichiometry

Stoichiometry involves using the quantitative relationships between reactants and products in a balanced chemical equation to determine amounts, such as mass or moles, of substances consumed or produced. It is essential for calculating yields, determining limiting reactants, and predicting the amounts of products formed in a chemical reaction.

Balanced Chemical Equations

Balanced chemical equations provide the foundation for stoichiometric calculations by showing the relative amounts of reactants and products involved in a chemical reaction. Each coefficient in the equation represents the mole ratio that must be maintained to satisfy the law of conservation of mass.

Limiting Reactant Analysis

Limiting reactant analysis is the process of determining which reactant will be completely consumed first in a chemical reaction, thereby limiting the amount of product formed. This analysis is crucial for accurately calculating yields and understanding the reaction dynamics.

Molar Mass Calculations

Molar mass calculations convert between the mass of a substance and the number of moles, providing a necessary link in stoichiometry. Knowing the molar masses allows one to relate experimental mass measurements to the mole ratios used in balancing chemical equations.

Combustion Reactions in Air

Combustion reactions in air typically involve the reaction of a substance with oxygen, but may also include reactions with other atmospheric components such as nitrogen. Understanding these reactions is important as they often yield multiple products, which requires multi-step stoichiometric analysis.

Indirect Analysis via Secondary Reactions

Indirect analysis involves determining the amount of a substance by measuring the quantity of a product generated from a subsequent reaction. Using information from secondary reactions, such as the hydrolysis of a compound producing a detectable gaseous product, allows for the back-calculation of the original amounts involved in the primary reaction.

Transcript

00:04 We have given reaction is 2 of m .g plus o 2 into 2m .m .g.

00:18 O 2m .o and 3 mol of m .g plus n2 into and m g and 2 and 2 and 2 and 2 and 3 3 into plus 6 mol of h2o into 3 mole of m .g .o .h.

00:52 Whole twice plus 2 mol of nh 3.

01:02 The mass of magnesium is equal to to 1 .496 gram.

01:21 And the mass of magnesium is equal to 1 .46 gram.

01:31 Of n .h3 is mass of nh3 is equal to 2 .813 gram.

01:50 We have to find the mass of m g3 and 2 and m g0 first we will find the number of moles of number of moles of number of moles and mass of mg3 and 2 and the molar mass of nh3 is we have molar mass of nh3 is equal to molar mass of nitrogen plus 3 into molar mass of hydrogen so by putting the molar mass of nitrogen 14 .007 gram per mole plus in three into molar mass of hydrogen is 1 .008 gram per mole is equal to 17 .031 gram per mole.

04:03 And the number of moles of nh3 is number of moles of nh3 is number of moles of nh3.

04:26 Is equal to mass of nh3 divided by molar mass of nh3.

04:37 Is equal to we have mass of nh3 is 2 .813 grams divided by molar mass of nh3 is 17 .031 per mole is equal to year gram cancel with gram.

05:01 And we have number of moles of nh3 is 0 .165 mole.

05:16 And since 1 mole of magnesium 3 and 2 produce 2 moles of nh3 0 .165 moles of nh3 0 .165 moles of nh3 will be produced by 0 .165 divided by 2 is equal to we have 8 .25 of into 10 raise to the power minus 2 moles of moles of mg 3 and 2.

06:52 So the molar mass of mg 3 and 2 is molar mass of the compound is equal to 3 into molar mass of magnesium plus 2 into molar mass of nitrogen is equal to 3 into molar mass of magnesium is 24 .305 gram per mole plus 2 into molar mass of nitrogen is 14 .007 gram per mole is equal to 100 .007 gram per mole is equal to 100 point 9 to 9 gram per mole.

08:13 Therefore the mass of the compound the compound m gn2 is mass of mj 3 and 2 is equal to number of moles of m g 3 and 2 into into molar mass of mj 3 and 2 is equal to 8 .25 into 10 raise to the power minus 2 mol into molar mass is 100 .9 to 9 gram per mole.

09:18 Is equal to we have mass of mg3 and 2 is 8 .3 to 7 to 7 gram.

09:33 Now we will find the mass of mg that react with n2...

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