Question

for the equation 6 HNO+2Fe->3H2+2Fe(NO3)3, 21.4 grams of HNO3 is reacted with 20.3 grams Fe. How...

for the equation 6 HNO+2Fe->3H2+2Fe(NO3)3, 21.4 grams of HNO3 is reacted with 20.3 grams Fe. How many grams of Fe(NO3)3 will form?

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Answer #1

6 HNO3+2Fe->3H2+2Fe(NO3)3,

no of moles of HNO3 =W/G.M.Wt

                                 = 21.4/63 = 0.339 moles

no of moles of Fe = W/G.A.Wt

                           = 20.3/56 = 0.3625moles

From balanced equation

2 moles of Fe react with 6 moles of HNO3

0.3625 moles of Fe react with = 6*0.3625/2 = 1.0875moles of HNO3

limiting reagent is HNO3

6 moles of HNO3 react with Fe to form 2 moles of Fe(NO3)3

6*63gm of HNO3 react with Fe to form 2*242 gm of Fe(NO3)3

21.4 gm of HNO3 react with Fe to form = 2*242*21.4/6*63 = 10357.6/378 = 27.4 gm of Fe(NO3)3

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