consider the reaction of 215ml 2.15M nitric acid with 26.0 g of solid iron(2) ca

ID: 780492 • Letter: C

Question

consider the reaction of 215ml 2.15M nitric acid with 26.0 g of solid iron(2) carbonate for the following questions

1. write the balanced chemical equation for this reaction

2. which reactant is the limiting reactnat?

3. what volume of carbon dioxide is produced at 755.8 torr and 22.3 'C?

4. determine the volume of carbon dioxide actually produced if the reaction completed with 65.6% yield?

5. what volume of water is created as a result of the reaction? density water=0.0997701 g/ml

6. Determine the concentration of iron(2) carbonate that is formed?

ans1)

FeCO3 + 4 HNO3 --> Fe(NO3)3 + CO2 + NO2 + 2 H2O

ans2)

moles of HNO3 = 0.215*2.15 = 0.46225 moles

moles of FeCO3 = 26/159.69 = 0.1628 moles

so 1 mole FeCO3 reacts with 4 mole HNO3

so 0.1628 moles FeCO3 reacts with 0.1628*4 = 0.6512 moles of HNO3

so limiting reagent is HNO3

ans3)

so as HNO3 is limiting 0.46225 moles moles HNO3 gives 1/4*0.46225 moles CO2

so moles of CO2 = 0.11556 moles CO2

so P = 755.8 torr = 0.994 atm

T = 295.3 K

R = 0.0821

V = nRT/P

V = 2.8185 L CO2

ans4)

if reaction gives 65.6% yeild

moles of CO2 = 0.656*0.11556 = 0.0758 moles

so V = 0.0758*0.0821*295.3/0.994

V = 1.8488 L CO2

ans5)

moles of H2O = 2/4*0.46225 = 0.231 moles H2O

so mass of H2O = 0.231*18 = 4.158 gm

volume = mass/density = 4.158/0.0997701

volume of H2O = 41.6758 ml H2O

ans6)

please clarify the question in the reaction iron(II) nitrate is formed not iron(II) carbonate

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