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I totally need this answer in 20 min. With solution huhuhuhu.......

1.Three (3.0) grams of sodium bicarbonate(NaHCO3) and 1.0 gram of citric acid(H3C6H5O7) are combined.Which is the limiting reactant,and what volume of CO2 will be produced?

2.What mass of lead(II) iodide will be formed when 20.0 g of lead(II) nitrate is added to 33.0 g of potassium iodide? Which is the limiting reactant?

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1 mole of NaHCO3 (base) reacts with 1 mole of citric acid to make 1 mole of CO2
basically HCO3- + H+ --> H20 + C02
3g of NaHCO3 is 3/81 moles
1g of citric acid is 1/155 moles = and being much smaller is the limiting reagent
therefore
only 1/155 moles of CO2 is produced
at room temp the volume will be 1/155 x 24.4 litres.

I'll try the other one now

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Pb(NO3)2 + 2KI ---> 2KNO3 + PbI2
n[Pb(NO3)2] = 22/(207.2+62x2)) = approx 0.067 moles <-- in short supply
n(KI) = 33/(39.1+126.9) = approx 0.200 moles
therefore
mass(KPbI2) formed will be 0.067 x (207.2 + 126.9x2) = approx 20g
[sorry haven't checked calulations)
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Thank you very very much dude! <3
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Hey!whats the answer on number 1 and 2 :)
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Final answers roughly (too zonked to use calculator)
1) 0.16litres
2) 20g
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