Answer :
[tex]\boxed{2.6264752{\text{ g}}}[/tex] of NO is formed during the given reaction.
Further explanation:
Limiting reagent is the species or reagent in a chemical reaction according to which the amount of product is decided. It is completely consumed in the reaction.
Stoichiometry governs the amount of species present in the reaction by using the relationship between reactants and products.
The given reaction is as follows:
[tex]{\text{4N}}{{\text{H}}_3} + {\text{5}}{{\text{O}}_2} \to {\text{4NO}} + 6{{\text{H}}_{\text{2}}}{\text{O}}[/tex]
The formula to calculate the moles of the substance is as follows:
[tex]{\text{Moles of substance}} = \dfrac{{{\text{Given mass of substance}}}}{{{\text{Molar mass of substance}}}}[/tex] ...... (1)
Substitute 3.25 g for the given mass and 17.03 g/mol for the molar mass in equation (1) to calculate the moles of [tex]\text{NH}_3[/tex].
[tex]\begin{aligned}{\text{Moles of N}}{{\text{H}}_3} &= \left( {\frac{{{\text{3}}{\text{.25 g}}}}{{17.03{\text{ g/mol}}}}} \right)\\&= 0.1908{\text{ mol}}\\\end{aligned}[/tex]
Substitute 3.50 g for the given mass and 31.98 g/mol for the molar mass in equation (1) to calculate the moles of [tex]\text{O}_2[/tex].
[tex]\begin{aligned}{\text{Moles of }}{{\text{O}}_2} &= \left( {\frac{{{\text{3}}{\text{.50 g}}}}{{{\text{31}}{\text{.98 g/mol}}}}} \right)\\&= 0.1094{\text{ mol}}\\\end{aligned}[/tex]
According to the stoichiometry of this reaction, four moles of reacts with five moles of [tex]\text{O}_2[/tex] to form four moles of NO and six moles of [tex]\text{H}_2\text{O}[/tex].
Calculate the moles of [tex]\text{O}_2[/tex] that react with and 0.1908 moles of [tex]\text{NH}_3[/tex] as follows:
[tex]\begin{aligned}{\text{Amount of }}{{\text{O}}_2} &= \left( {{\text{0}}{\text{.1908 mol N}}{{\text{H}}_3}} \right)\left( {\frac{{{\text{5 mol }}{{\text{O}}_2}}}{{4{\text{ mol N}}{{\text{H}}_3}}}} \right)\\&= 0.2385{\text{ mol }}{{\text{O}}_2}\\\end{aligned}[/tex]
The above calculations concluded that 0.2385 moles of [tex]\text{O}_2[/tex] are needed to react with 0.1908 moles of [tex]\text{NH}_3[/tex] but only 0.1094 moles of [tex]\text{O}_2[/tex] are present in the reaction mixture. Therefore, [tex]\text{O}_2[/tex] is present in limited quantity and is a limiting reagent.
Since [tex]\text{O}_2[/tex] comes out to be the limiting reagent, the formation of NO is governed by it. According to the reaction stoichiometry, five moles of [tex]\text{O}_2[/tex] produces four moles of NO. Therefore the amount of NO can be calculated as follows:
[tex]\begin{aligned}{\text{Amount of NO}} &= \left( {0.1094{\text{ mol }}{{\text{O}}_2}} \right)\left( {\frac{{4{\text{ mol NO}}}}{{5{\text{ mol }}{{\text{O}}_2}}}} \right)\\&= 0.08752{\text{ mol}}\\\end{aligned}[/tex]
Rearrange equation (1) to calculate the mass of the substance.
[tex]{\text{Mass of substance}}= \left( {{\text{Moles of substance}}} \right)\left( {{\text{Molar mass of substance}}} \right)[/tex] ...... (2)
Substitute 0.08752 mol for the moles and 30.01 g/mol for the molar mass in equation (2) to calculate the mass of NO.
[tex]\begin{aligned}{\text{Mass of NO}} &= \left( {0.08752{\text{ mol}}} \right)\left( {30.01{\text{ g/mol}}} \right)\\&= 2.6264752{\text{ g}}\\\end{aligned}[/tex]
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Answer details:
Grade: Senior School
Chapter: Mole concept
Subject: Chemistry
Keywords: limiting reagent, NH3, NO, H2O, O2, 2.6264752 g, moles, mass, molar mass.