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23 November, 09:49

You carefully weigh out 13.00 g of CaCO3 powder and add it to 52.65 g of HCl solution. You notice bubbles as a reaction takes place. You then weigh the resulting solution and find that it has a mass of 60.32

g. The relevant equation isCaCO3 (s) + 2HCl (aq) ? H2O (l) + CO2 (

g.+CaCl2 (aq) Assuming no other reactions take place, what mass of CO2 was produced in this reaction? Express your answer to three significant figures and include the appropriate units.

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  1. 23 November, 11:43
    0
    CaCO₃ + 2HCl = CaCl₂ + CO₂ + H₂O

    n (CaCO₃) = m (CaCO₃) / M (CaCO₃)

    n (CaCO₃) = 13.00/100.09=0.1299 mol

    Δm=13.00+52.65-60.32=5.33 g

    m (CO₂) = 5.33 g

    n (CO₂) = 5.33/44.01=0,1211 mol

    w=0.1211/0.1299=0,9323 (93.23%)
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