Asked by hbl

Calcium carbonate reacts with HCl according to the following equation:
2HCl(aq)+CaCO3(s)→CaCl2(aq)+H2O(l)+CO2(g)
How many moles of HCl are in 58 mL of 0.13 M HCl?
What mass of calcium carbonate is needed for complete reaction with the HCl in (a)?
Answered by bobpursley
molesHCl=.058*.13
you need mole of calcium carbonate 1/2 the moles of HCl
Answered by hbl
that is NOT the correct answer for the mass of calcium carbonate needed for complete reaction with the HCl in part (a) I tried the same thing and a few other ways and all are wrong---any additional suggestions
Answered by DrBob222
hbl, I think you jumped the gun. mols CaCO3 = 1/2 mols HCl as Bob P wrote. He assumed you would be able to convert mols to grams.
Then mass CaCO3 = mols CaCO3 x molar mass CaCO3 = ?
Answered by Bob
Deez nutz 069
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