{"id":432150,"date":"2018-03-29T12:58:33","date_gmt":"2018-03-29T12:58:33","guid":{"rendered":"https:\/\/essaypaper.org\/easy-calculation-of-reactant-and-product-mass-with-expert-solution\/"},"modified":"2018-10-24T08:57:05","modified_gmt":"2018-10-24T08:57:05","slug":"easy-calculation-of-reactant-and-product-mass-with-expert-solution","status":"publish","type":"post","link":"https:\/\/www.benedictsol.com\/blogs\/easy-calculation-of-reactant-and-product-mass-with-expert-solution\/","title":{"rendered":"Easy Calculation of Reactant and Product Mass with Expert Solution"},"content":{"rendered":"\n
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Calculation of reacting masses is a difficult aspect altogether as it involves using a balanced simple equation. Apart from that it also needs one to have proper knowledge of moles and masses.<\/p>\n

A simple balanced equation will look something like this,<\/p>\n

2Mg + O2\u00a0 <\/sub>\u00e0 2MgO<\/p>\n

So, taking the above equation in consideration, let us calculate how much magnesium oxide is produced from 60 g of Mg?<\/p>\n

The difficult part of its calculation is the difference in mass of Mg in reactant and product respectively as the relative mass of magnesium oxide and magnesium is different.<\/p>\n

Careful calculation and after rearranging the equation for mass equally number of moles times by the relative mass the answer is 100g.<\/p>\n

The components involved in calculation of the mass are relative mass, moles, and atomic mass unit that will be the same as that of the reactants taking part in the reaction.<\/p>\n

Another tricky question is as follows \u2013<\/strong><\/p>\n

What mass of PCl5 will be produced from the given masses of both reactants? <\/strong>Mass of P4 is 26.0 g and excess Cl2.<\/p>\n

The solution is a bit elaborated.<\/p>\n

Since, we have, P4 <\/sub>+ 10 Cl2 <\/sub>= 4 PCl5<\/sub><\/p>\n

Addition of 1 mole of P4 <\/sub>with 10 moles of Cl2 <\/sub>will give 4 moles of PCl5 <\/sub><\/p>\n

Hence, molecular weight of P4 <\/sub>= 123.895048 gmol-1<\/sup><\/p>\n

Molecular weight of Cl2 <\/sub>= 70.906 gmol-1<\/sup><\/p>\n

Molecular weight of PCl5 <\/sub>= 208.22 gmol-1<\/sup><\/p>\n

The next part will be something like this,<\/p>\n

Number of P4 <\/sub>= mass of P4 <\/sub>\/ molecular weight of P4 <\/sub>= 26.0 g \/ 123.895048 gmol-1 <\/sup>= 0.210 mol<\/p>\n

Number of PCl5 <\/sup>= 4 number of P4 <\/sub>= 0.839 mol<\/p>\n

And the second last part is,<\/p>\n

Number of Cl2 <\/sub>= mass of Cl2 <\/sub>\/ molecular weight of Cl2<\/sub> = 59.0 g \/ 70.906 gmol-1<\/sup> = 0.832 mol<\/p>\n

Number of Cl5 <\/sub>= 0.4 number of Cl2<\/sub> = 0.333 mol<\/p>\n

Last part of the calculation is,<\/p>\n

Here chlorine is limiting reagent.<\/p>\n

0.333 moles of PCl5<\/sub> will be formed.<\/p>\n

Mass of PCl5<\/sub> = number of PCl5<\/sub> x molecular weight of PCl5<\/sub> = 0.333 mol x 208.22 gmol-1<\/sup> = 69.3 g<\/p>\n

Same like that of the question what mass of PCl5 will be produced from the given masses of both reactants? <\/strong><\/p>\n

The easiest steps to find out reacting masses are by setting up an equation and then balancing it. Both parts B and C have limiting reagents that provides the theoretical yield upon further calculation.<\/p>\n

Taking another question into consideration,<\/p>\n

    \n
  1. Q) What is the mass of C02 <\/sub>formed from 4 g of methane?<\/strong><\/li>\n<\/ol>\n

    First of all the equation will be,<\/p>\n

    CH4 <\/sub>+ 2O2 <\/sub>\u00e0 CO2<\/sub> + 2H2<\/sub>O<\/p>\n

    One method to answer this question is to use reacting masses that can be done by writing down the formula masses for everything present in the equation including multiplying for the number of something present in the equation.<\/p>\n

    Calculating the reactant and product masses can be made easier with the small tips, only the basic knowledge about the molecular formula and masses is something a student need to remember.<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

    Calculation of reacting masses is a difficult aspect altogether as it involves using a balanced simple equation. Apart from that it also needs one to have proper knowledge of moles and masses. A simple balanced equation will look something like Read More …<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[99,17,41],"tags":[],"class_list":["post-432150","post","type-post","status-publish","format-standard","hentry","category-academic-essays","category-academic-assignment-help","category-academic-essay-writing-uk"],"_links":{"self":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts\/432150","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/comments?post=432150"}],"version-history":[{"count":0,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts\/432150\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/media?parent=432150"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/categories?post=432150"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/tags?post=432150"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}