anonymous
  • anonymous
Okay I need help with this question. Consider the production of phosphoric acid. ___P4O10+___H20 ---> ___HPO4 What is the theoretical yield of phosphoric acid if 210. grams of tetraphosphorous decoxide is reacted with excess water?
Chemistry
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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aaronq
  • aaronq
This is pretty easy, i'll give you the outline, read it and try it. I'll help you fix it if you make mistakes. Stoichiometry When performing stoichiometric calculations the values to be used \(\sf \color{red}{must~be~in~moles}\)*. This is because the Molar Masses (\(\sf M_{m}\) of two elements are not the same. This is to say that 1 gram of Hydrogen atoms is not the same number of atoms as 1 gram of Nitrogen atoms. We need to be working with numbers of particles (e.g. atoms, molecules, formula units) and using moles is the way this is done. *A shortcut can be taken when working with \(\sf \color{blue}{Ideal~Gases}\) as moles are \(\propto\) volume. And, so volume can be regarded as moles, where 22.414 L = 1 mol. General Scheme: \(\sf \large 1.\)First write and balance an equation for the process described. \(\sf \large 2.\)Next, use the stoichiometric coefficients to find moles produced. \(\sf \large 3.\) Set up a ratio using the species of interest, like so (for a general reaction): \(\sf \large \color{red}{a}A + \color{blue}{b}B \rightleftharpoons \color{green}{c}C\) here upper case letters (A,B,C) are the chemical species, and lower case letters (a,b,c) are the coefficients, \(\sf \dfrac{moles_A}{\color{red}{a}}=\dfrac{moles_B}{\color{blue}{b}}=\dfrac{moles_C}{\color{green}{c}}\) You can use any pair from here, depending on what the question is asking for. For example: if you have 2 moles of B, how many moles of C can you produce? Now rearrange algebraically, \(\sf\dfrac{2}{\color{blue}{b}}=\dfrac{moles _C}{\color{green}{c}}\rightarrow moles _C=\dfrac{2*\color{green}{c}}{\color{blue}{b}}\) \(\sf \large 4.\) Solve ------------ You'll also need this formula to find moles: \(\sf moles=\dfrac{mass}{Molar~mass}\)

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