Look at the equation:
each mole of CaCO3 produces one mole of CaCl2.
So, you will wind up with .3097 moles of CaCl2.
Convert that to grams, and you're done.
CaCO3(s)+2HCl(aq)-->CaCl2(aq)+H2O(l)+CO2 (g)
How many grams of calcium chloride will be produced when 31g of calcium carbonate are combined with 11 g of HCl?
I have 0.3097 moles of CaCO3
and 0.30169 moles of HCl...I get stuck at that point
2 answers
CaCO3(s)+2HCl(aq)-->CaCl2(aq)+H2O(l)+CO2 (g)
0.3097 moles of CaCO3
0.30169 moles of HCl
0.3097 Moles CaCO3 X (1 Mole CaCl2 / 1 Mole CaCO3) = 0.3097 Mole CaCL2
0.30169 Moles HCL X ( 1 Mole CaCl2 / 2 Mole HCL) = 0.1508 Mole CaCl2
I think HCL is a limiting reactant: Therefore: 0.1508 Mole CaCl2 is the product amount.
0.3097 moles of CaCO3
0.30169 moles of HCl
0.3097 Moles CaCO3 X (1 Mole CaCl2 / 1 Mole CaCO3) = 0.3097 Mole CaCL2
0.30169 Moles HCL X ( 1 Mole CaCl2 / 2 Mole HCL) = 0.1508 Mole CaCl2
I think HCL is a limiting reactant: Therefore: 0.1508 Mole CaCl2 is the product amount.
1 answer