Asked by bc
You carefully weigh out 20.00g of \rm CaCO_3 powder and add it to 81.00g of \rm HCl solution. You notice bubbles as a reaction takes place. You then weigh the resulting solution and find that it has a mass of 92.80g . The relevant equation is
\rm CaCO_3{\it (s)}+2HCl{\it (aq)}\rightarrow H_2O{\it (l)}+CO_2{\it (g)}+CaCl_2{\it (aq)}
Assuming no other reactions take place, what mass of \rm CO_2 was produced in this reaction?
\rm CaCO_3{\it (s)}+2HCl{\it (aq)}\rightarrow H_2O{\it (l)}+CO_2{\it (g)}+CaCl_2{\it (aq)}
Assuming no other reactions take place, what mass of \rm CO_2 was produced in this reaction?
Answers
Answered by
DrBob222
Something about this problem doesn't add up.
If you go with 20 g CaCO3, that is 20/100 = 0.2 mol and that will produce 0.2 x 44 = 8.80g CO2.
However, if you add 81.00 g to 20 g to make 101 g total and you reweigh after the CO2 is expelled, the loss in mass is 101.0-92.80 = 8.20g.
It seems to me that the answers should be the same no matter which way the problem is worked.
If you go with 20 g CaCO3, that is 20/100 = 0.2 mol and that will produce 0.2 x 44 = 8.80g CO2.
However, if you add 81.00 g to 20 g to make 101 g total and you reweigh after the CO2 is expelled, the loss in mass is 101.0-92.80 = 8.20g.
It seems to me that the answers should be the same no matter which way the problem is worked.
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