F4 Chem HELP!!!! 急

[匿名]

2013-02-24 5:07 am

1. A 300cm3 flask is filled with carbon dioxide at room temperature and pressure. How many cabon dioxide molecules are there in the falsk?

2. If a mixture of 0.80g of oxyen and 0.20g of hydrogen is exploded,what will be the volume of remaining gas at room temperture and pressure?

3.Sodium hydrogencarbonate decomposes upon heating to give sodium carbonate, carbon dioxide and water. What volume of cabon dioxide at room temperture and atmospgeric pressure can be obtained by decomposing 1.00g of sodium hydrohencarbonate?

4.Calcium carbonate can be attacked by an acid.A sample of calcium carbonate is mixed with excess hydrochric acid, the volume of carbon dioxide formed at room temperture and pressure is 180com3. Find the mass of calcium carbonate in the sample.

5.A sample of limestone caontains mainly calcium carbonate.1.00g of the sample was heated,210cm3 of cabon dioxide is collected at room temperture and pressure.Find the percentage by mass of pure calcium carbonate in the sample.

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回答 (1)

不用客氣

2013-02-24 6:09 am

✔ 最佳答案

1.
No. of moles of CO2 = 300/24000 = 0.0125 mol
No. of CO2 molecules = 0.0125 x (6.02 x 10²³) = 7.525 x 10²¹

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2.
2H2(g) + O2(g) → 2H2O(l)

No. of moles of 0.20 g of H2 = 0.20/(1x2) = 0.1 mol
No. of moles of 0.80 g of O2 = 0.80/(16x2) = 0.025 mol

For complete reaction of 0.025 mol of O2,
the no. of moles of H2 needed = 0.025 x 2 = 0.05 mol < 0.1 mol
Hence, H2 is in excess and O2 is the limiting reactant(completed reacted).

No. of moles of H2 gas left = 0 mol
No. of moles of O2 gas left = 0.1 - 0.05 = 0.05 mol
No. of moles of H2O gas formed ≈ 0 mol
Volume of remaining gas = (0 + 0.05 + 0) x 24000 = 1200 cm³

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3.
2NaHCO3 → Na2CO3+ H2O + CO2

No. of moles of NaHCO3 decomposed = 1.00/(23 + 1 + 12 + 16x3) =1.00/84 mol
No. of moles of CO2 formed = (1.00/84) x (1/2) = 1.00/168 mol
Volume of CO2 formed = (1.00/168) x 24000 = 143 cm³

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4.
CaCO3 + 2HCl → CaCl2 + H2O+ CO2

No. of moles of CO2 = 180/24000 = 0.0075 mol
No. of moles of CaCO3 = 0.0075 mol
Mass of CaCO3 in the sample = 0.0075 x (40 + 12 + 16x3) = 0.75 g

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5.
CaCO3 → CaO + CO2

No. of moles of CO2 = 210/24000 = 0.00875 mol
No. of moles of CaCO3 = 0.00875mol
Mass of CaCO3 = 0.00875 x (40 + 12 + 16x3) = 0.875 g
% by mass of CaCO3 = (0.875/1.00) x 100% = 87.5%

參考: 不用客氣

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