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1) From the data is the ratio of metal to hydrogen gas 2:1, 1:1, or 2:3? Briefly

ID: 1048325 • Letter: 1

Question

1) From the data is the ratio of metal to hydrogen gas 2:1, 1:1, or 2:3? Briefly explain.

2) The equation for the trendline follows the formula y = mx + b, where b is the y-intercept. The y-intercept should be set to zero. Briefly explain why.

3) If the water temperature was 30.0C and the atmospheric pressure was 753 torr. What volume of hydrogen gas would you expect to collect if you reacted 0.083g of Ca metal with HCl??

Trial 1 0.046 0.0018 Trial 2 0.056 0.0023 Trial 4 0.075 0.0031 Trial 5 0.087 0.0036 Trial 6 0.093 0.0038 Trial 3 0.066 1.Mass of Metal used (g) [2].Moles of Metal (mol) 3. Temperature of water in Celsius (C) [4] Temperature of water in Kelvin (K) 5. Volume of hydrogen gas (mL) [6] Volume of hydrogen gas (L) [7] Water vapor pressure (torr) 8. Atmospheric Pressure (torr) [9] Corrected H2 Pressure (torr) [10] Moles of Hydrogen Gas collected (mol) 0.0027 21 21 21 21 21 21 294.15 20 0.020 18.65 294.15 294.15 294.15 294.15 60 0.060 294.15 40 59 70 0.033 0.040 0.059 0.070 18.65 18.65 18.65 18.65 18.65 759.2 740.55 740.55 740.55 740.55 740.55 740.55 0.0008 0.0013 0.001610.0023 0.00240.00282 Data to Graph Y data 0.0008 0.0013 Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 Trial 6 X data 0.0018 0.0023 0.00270.00161 0.0031 0.0036 0.0038 0.00282 0.0023 0.0024

Explanation / Answer

From the given data,

averaging out moles of metal and moles of H2 of collected gives a ratio,

moles(meta/H2) = 2

So the metal to hydrogen gas ratio of 2 : 1

So for every 2 moles of metal reacting, we form 1 mole of H2 gas.

2) No H2 gas evolved when no HCl was present in the reaction, which gives a intercept at y-axis a value of zero.

3) moles of Ca = 0.083/40.08 = 0.0021 mols

moles of H2 = 0.0021 mols

volume of H2 collected = nRT/P

P = 753 - 31.8 = 721.2/760 = 0.95 atm

So,

volume of H2 collected = 0.0021 x 0.08205 x (273 + 30)/0.95 = 0.055 L