With the equation <span>A₂+2B₂→2AB₂ we know that one mole of A₂ is necessary to react with two moles of B₂ and with that, we obtain two moles of the product. ( you can see this by just seeing what numbers are before the letters)
Since it's needed more quantity of B than A, B is the limited reagent.
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By doing fractional distillation
<span>1 mole of calcium carbonate reacts with 1 mole of sulfuric acid and produces 1 mole of calcium sulfate.
3.1660 g of CaCO3 is how many moles of calcium carbonate? 3.1660 / 100.0869 = 0.031633 moles.
3.2900 g of H2S04 is how many moles of sulfuric acid? 3.2900 / 98.079 = 0.033544 moles.
</span><span>The lesser of the two is 0.031633 moles.
Therefore, 0.031633 moles of calcium carbonate will combine with 0.031633 moles of sulfuric acid to produce 0.031633 moles of calcium sulfate.
Molecular weight of calcium sulfate is 136.14 g/mol.
Therefore, 0.031633 moles of calcium sulfate will weight 0.031633 x 136.14 g/mol = 4.3065 grams.</span>
Answer:
0.924 g
Explanation:
The following data were obtained from the question:
Volume of CO2 at RTP = 0.50 dm³
Mass of CO2 =?
Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:
1 mole of gas = 24 dm³ at RTP
Thus,
1 mole of CO2 occupies 24 dm³ at RTP.
Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e
Xmol of CO2 = 0.5 /24
Xmol of CO2 = 0.021 mole
Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.
Finally, we shall determine the mass of CO2 as follow:
Mole of CO2 = 0.021 mole
Molar mass of CO2 = 12 + (2×16) = 13 + 32 = 44 g/mol
Mass of CO2 =?
Mole = mass /Molar mass
0.021 = mass of CO2 /44
Cross multiply
Mass of CO2 = 0.021 × 44
Mass of CO2 = 0.924 g.
I think you add 29.57 + 80 and the answer would be 30.37
