Assume x grams of CaCO3, so 7.85-x grams of MgCO3.
Now, you can figure the moles of each (in terms of X), and then the moles of CO2 (in terms of x). Yes, it is messy algebra, but you add those moles and set equal to .0817, and you can solve for x. Double check each line of algebra.
A mixture contained CaCO3 and MgCo3. A sample of this mixture weighing 7.85g was reacted with excess HCl. The reactions are--
CaCO3 + 2HCl --> CaCl2 + H2O + CO2
MgCO3 + 2HCl --> MgCl2 + H2O + CO2
Is the sample reacted completely and produced 1.94L of CO2, at 25 degrees C, 785mmhg, whre were the percentages of CaCO3 and MgCO3 in the mixture?
So far, I got the moles of CO2 by using PV=nRT, which is .0817moles of CO2,
I don't kno what to do next.. can anyone PLEASE help!
2 answers
Ca Co3(s) 2HCl (aq) Ca Cl2(aq)+H2o(e)+Co2(g)
2 answers