Respuesta :
Na2CO3 + 2Cl- ⇒ 2NaCl + CO3^-2
1 mole of Na2CO3 = 106 g
2 moles of NaCl = 2 x 58.4
= 116.8 g
Na2CO3 would increase by 116.8 / 106 = 1.10 to form 2NaCl.
0.4862 g x 1.10 = 0.515 grams of NaCl.
K2CO3 + 2Cl- ⇒ 2KCl + CO3^-2
1 mole of K2CO3 = 138.2 g
2 moles of KCl = 149.1
K2CO3 would increase by 149.1 /138.2 = 1.079 to form 2KCl
0.4862 x 1.079 = 0.5246 g
1 mole of Na2CO3 = 106 g
2 moles of NaCl = 2 x 58.4
= 116.8 g
Na2CO3 would increase by 116.8 / 106 = 1.10 to form 2NaCl.
0.4862 g x 1.10 = 0.515 grams of NaCl.
K2CO3 + 2Cl- ⇒ 2KCl + CO3^-2
1 mole of K2CO3 = 138.2 g
2 moles of KCl = 149.1
K2CO3 would increase by 149.1 /138.2 = 1.079 to form 2KCl
0.4862 x 1.079 = 0.5246 g
Answer:
The sample is a carbonate.
The unknown samples are Na2CO3 and K2CO3.
Explanation:
Let's assume ,the sample is Na2CO3.
Na2CO3 produces 2 moles of NaCl via formation of Na2O
Number of moles of NaCO3= 0.48262g/106gmol= 4.586×10^-3mol
Mass of NaCl= 2(4.586×10^-3)×58.5=0.536g
Let's assume the sample is K2CO3
K2CO3 produces2 moles of KCl
Number of moles of K2CO3= 0.4862g/138.21gmol
Number of moles= 3.517×10^-3mol
Mass of KCl formed= 2(3.517×10^-3)×74.55=0.5246g