Question
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
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
Answers
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.
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)
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.