Respuesta :
Ammonium carbonate is a salt that is soluble and is made up of ammonium radicle and a carbonate radicle. If the salt ionizes;
(NH4)2CO3 = 2NH4+ + CO3^-2, this means 1 mole of the salt contains two moles of ammonium ions.
1 mole of ammonium carbonate contains 96.0858 g
Therefore, 6.945 g are equivalent to (6.945/96.0858) = 0.07228 moles
Thus, 6.945 g of ammonium carbonate contains 2 × 0.07228 = 0.14456 moles.
(NH4)2CO3 = 2NH4+ + CO3^-2, this means 1 mole of the salt contains two moles of ammonium ions.
1 mole of ammonium carbonate contains 96.0858 g
Therefore, 6.945 g are equivalent to (6.945/96.0858) = 0.07228 moles
Thus, 6.945 g of ammonium carbonate contains 2 × 0.07228 = 0.14456 moles.
[tex]\boxed{0.14456{\text{ mol}}}[/tex] of ammonium ions are present in 6.945 g of ammonium carbonate.
Further explanation:
Stoichiometry
The amount of various species that are present in the chemical reaction is determined by stoichiometry. This is done by studying of relationship between reactants and products. It is used to determine moles of a chemical species when moles of other chemical species in the reaction is given.
The general chemical equation is as follows:
[tex]{\text{X}} + 2{\text{Y}} \to 3{\text{Z}}[/tex]
Here,
X and Y are reactants.
Z is the product.
According to the above equation, one mole of X reacts with two moles of Y and as a result of the reaction between the both, three moles of Z are produced. Therefore the stoichiometric ratio between X and Y is 1:2, that between X and Z is 1:3 and that between Y and Z is 2:3.
The given reaction occurs as follows:
[tex]{\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} \rightleftharpoons 2{\text{NH}}_4^ + + {\text{CO}}_3^{2 - }[/tex]
The formula to calculate the moles of [tex]{\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] is as follows:
[tex]{\text{Moles of}}{\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} = \dfrac{{{\text{Given mass of }}{{\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}}{{{\text{Molar mass of }}{{\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}}[/tex] …… (1)
Substitute 6.945 g for the given mass and 96.09 g/mol for the molar mass of in equation (1).
[tex]\begin{aligned}{\text{Moles of}}{\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} &= \left( {6.945{\text{ g}}} \right)\left({\frac{{1{\text{ mol}}}}{{96.09{\text{ g}}}}} \right)\\&= 0.07228{\text{ mol}}\\\end{aligned}[/tex]
According to the reaction stoichiometry, one mole of ammonium carbonate forms two moles of ammonium ion and one mole of carbonate ion.
The formula to calculate the moles of ammonium ions is as follows:
[tex]{\text{Moles of NH}}_4^ + = 2\left( {{\text{Moles of }}{{\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right)[/tex] …… (2)
Substitute 0.07228 mol for the moles of [tex]{\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] in equation (2).
[tex]\begin{aligned}{\text{Moles of NH}}_4^ + &= 2\left( {{\text{0}}{\text{.07228 mol}}}\right)\\&= 0.14456{\text{ mol}}\\\end{aligned}[/tex]
Learn more:
- Calculate the moles of chlorine in 8 moles of carbon tetrachloride: https://brainly.com/question/3064603
- Calculate the moles of ions in the solution: https://brainly.com/question/5950133
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Mole concept
Keywords: stoichiometry, X, Z, Y, (NH4)2CO3, NH4+, CO32-, moles, 0.07228 mol, 0.14456 mol, reactants, products.
