Question

plss answer this !!( mole concept and stoichiometry)

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Answer 1

I ll let you know the three formulae....those mole concept formulae can be applied anywhere:

1) no. Of moles=volume given /22.L(or 22.4dm cube)

2) no. Of moles =mass given/ atomic mass or molecular mass(here u can apply in the denominator either atomic mass if they ask mass of an atom or u can also keep molecular mass if they ask mass of a molecule)

3) no. Of moles=molecular mass or atomic mass/6.022×10 to the power 23.

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asopachitrita

Secondary SchoolChemistry 13+7 pts

2Ca(NO3)2 --- 2CaO + 4NO2 + O2

1) How many moles of NO2 are produced when 1 mole of Ca(NO3)2 decomposes ?

2) What volume of O2 at S.T.P. will be produced on heating 65.6g of Ca(NO3)2 ?

3) Find out the mass of CaOformed when 65.6g of Ca(NO3)2 is heated ?

4) Find out the mass of Ca(NO3)2required to produce 5 moles of gaseous products ?

5) Find out yhe mass of Ca(NO3)2 required to produce 44.8L of NO2 at S.T.P.

Report by Svsriramam6856 12.02.2018

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asopachitrita

Asopachitrita · Expert

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Nate3306 Helping Hand

1) - given the stoichiometric coefficients, we know that 2 moles of Ca(NO3)2 will produce 4 moles of NO2, hence, 1 mole will produce 2 moles of NO2

2) - 328 g produces 22.4 L, since One mole of any gas at S.T.P. occupies the same volume which is 22.4 L.

Hence, 65.6 g produces (65.6*22.4)/328 = 4.48 L

3) - 328 g produces 112 g CaO. Therefore, 65.6 g produces = (65.6*112)/328 = 22.4 g CaO.

4) - Given the stoichiometric coefficients, we know 5 moles of gaseous products are already being produces ( 4+1) by 2 moles of reactant.

5) - 44.8 L at STP = 2 moles of NO2, since One mole of any gas at S.T.P. occupies the same volume which is 22.4 L.

Hence, to produce 2 moles of NO2, we need 1 mole of reactant = 164 g