Acid precipitation dripping on Limestone produces Carbon Dioxide by the following reaction: CaCo3(s) + 2H+(aq) => Ca(2+)(aq) + CO2(g

Question

Acid precipitation dripping on Limestone produces Carbon Dioxide by the following reaction:
CaCo3(s) + 2H+(aq) —> Ca(2+)(aq) + CO2(g) + H2O (l)
15.5mL of CO2 was produced at 25*C and 738.0 mmHg
How man moles of CO2 were produced?
How many milligrams of CaCO3 were consumed?

in progress 0
2022-04-29T22:35:04+00:00 2 Answers 0

Answers ( 2 )

  1. Ama12a
    0
    2022-04-29T22:36:05+00:00

    1) PV=nRT
    P=738.0 mmHg
    V=15.5mL=0.0155 L
    T=273+25=298 K
    R=62.36 L*mmHg*Ku207bu00b9molu207bu00b9
    n=PV/RT
    n=(738.0mmHg *0.0155 L)/(62.36 L*mmHg*Ku207bu00b9molu207bu00b9*298 K)= =0.000616=6.16*10u207bu2074 mol

    2) From the equation of the reaction
    1 mol CaCO3 gives 1 mol CO2,
    so 6.16*10u207bu2074 mol CaCO3 —-> 6.16*10u207bu2074 mol CO2

    Molar mass CaCO3 =M(Ca)+M(C)+3*M(O)= 40.1+12.0+3*16.0 =100.1 g/mol
    6.16*10u207bu2074 mol CaCO3 * 100.1 g/mol =617*10u207bu2074 g =0.0617 g = 61.7mg

  2. Alice
    0
    2022-04-29T22:36:44+00:00

    Answer: The number of moles of carbon dioxide produced are [tex]6.16times 10^{-4}mol[/tex] and the mass of calcium carbonate is 61.6 mg

    Explanation:

    To calculate the moles of carbon dioxide gas, we use the equation given by ideal gas which follows:

    [tex]PV=nRT[/tex]

    where,

    P = pressure of the gas = 738.0 mmHg

    V = Volume of the gas = 15.5 mL = 0.0155 L (Conversion factor: 1 L = 1000 mL)

    T = Temperature of the gas = [tex]25^oC=[25+273]K=298K[/tex]

    R = Gas constant = [tex]62.3637text{ L.mmHg }mol^{-1}K^{-1}[/tex]

    n = number of moles of carbon dioxide gas = ?

    Putting values in above equation, we get:

    [tex]738.0mmHgtimes 0.0155L=ntimes 62.3637text{ L.mmHg }mol^{-1}K^{-1}times 298Kn=frac{738times 0.0155}{62.3637times 298}=6.16times 10^{-4}mol[/tex]

    For the given chemical equation:

    [tex]CaCO_3(s)+2H^+(aq.)rightarrow Ca^{2+}(aq.)+CO_2(g)+H_2O(l)[/tex]

    By Stoichiometry of the reaction:

    1 mole of carbon dioxide gas is produced from 1 mole of calcium carbonate

    So, [tex]6.16times 10^{-4}mol[/tex] of carbon dioxide will be produced from = [tex]frac{1}{1}times 6.16times 10^{-4}=6.16times 10^{-4}mol[/tex] of calcium carbonate

    • To calculate the mass for given number of moles, we use the equation:

    [tex]text{Number of moles}=frac{text{Given mass}}{text{Molar mass}}[/tex]

    Molar mass of calcium carbonate = 100 g/mol

    Moles of calcium carbonate = [tex]6.16times 10^{-4}mol[/tex]

    Putting values in above equation:

    [tex]6.16times 10^{-4}mol=frac{text{Mass of calcium carbonate}}{100g/mol}text{Mass of calcium carbonate}=(6.16times 10^{-4}moltimes 100g/mol)=6.16times 10^{-2}g[/tex]

    Converting this into milligrams, we use the conversion factor:

    1 g = 1000 mg

    So, [tex]6.16times 10^{-2}gtimes frac{1000mg}{1g}=61.6mg[/tex]

    Hence, the number of moles of carbon dioxide produced are [tex]6.16times 10^{-4}mol[/tex] and the mass of calcium carbonate is 61.6 mg

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