what is answer part A to part C? Part A For the reaction 4A1 (s) + 302 (g) 2A12O

Question

what is answer part A to part C?

Part A For the reaction 4A1 (s) + 302 (g) 2A12Og (s). What is the mass of Al2O3 that is produced when 22.85 g of Al and 22.85 g of O2 react? Submit My Answers Give Up Part B For the reaction 3NO2 (g) + H2O(l) 2HNOg (aq) + NO(g), What is the mass of HNO3 that is produced when 25.40 g of NO2 and 25.40 g of H2O react? Submit My Answers Give Up Part C For the reaction 4NH3 (g) + 502 (g) 4N0(g) + 6H20(g). What is the mass of H2O that is produced when 20.75 g of NH3 and 20.75 g of O2 react?

Explanation / Answer

part A

According to equation, 4 moles of Al(108g) reacts with 3moles of O2(96g

Then 22.85g of Al requires 96*22.85/108 =20.32g of O2 but O2 given is 22.85g which is in excess ,hence Al is limiting reagent

As per the equation 108g of Al forms 204g of Al2O3

Then 22.85g forms, 204*22.85/108=43.162g of Al2O3 is formed

PartB, as per equation 138g of NO2 combines with 18g of H2O

TheThen 25.4 g of N2O requires 18*25.4/138=3.31g of H2O but given H2O is 25.4g which is in excess, so NO2 is limiting reagent

As per equation 138g forms 126g of HNO3

Then 25.4g forms =126*25.4/138=23.19g of HNO3

PartC

As per equation 68g of NH3 reacts with 160g of O2

Then 20.75g of NH3 require 160*20.75/68=48.82g of O2 but given O2 is less hence O2 is limiting reagent

As per equation 160g of O2 produce 108g of H2O

But 20.75g of O2 produce 108*20.75/160 =14g of H2O

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