Question #55746
What mass of calcium carbonate in grams will be produced from 1.67 mole of sodium carbonate
Expert's answer
Ca(OH)2(aq) + Na2CO3 (aq) = CaCO3(s) + 2NaOH(aq)
From the stoichiometry of this reaction it is obviously that 1 mole of sodium carbonate yields to 1 mole of calcium carbonate, so
n(CaCO3) = n(Na2CO3) = 1.67 mol
The relationship between the amount of substance and the mass is as follows:
m(CaCO3) = n(CaCO3) M(CaCO3)
m(CaCO3) = 1.67x100.09 = 167.15 g
Answer: 167.15 g
From the stoichiometry of this reaction it is obviously that 1 mole of sodium carbonate yields to 1 mole of calcium carbonate, so
n(CaCO3) = n(Na2CO3) = 1.67 mol
The relationship between the amount of substance and the mass is as follows:
m(CaCO3) = n(CaCO3) M(CaCO3)
m(CaCO3) = 1.67x100.09 = 167.15 g
Answer: 167.15 g
"assignmentexpert.com" is professional group of people in Math subjects! They did assignments in very high level of mathematical modelling in the best quality. Thanks a lot
#339914 on Dec 2023
#339914 on Dec 2023