Answer:
0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ are necessaries
Explanation:
Based on the reaction
CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃
1 mole of CaCl₂ reacts per mole of Na₂CO₃
calculate how many moles of CaCl2•2H2O are present in 1.50 g...
To solve this question we must find the moles of CaCl2•2H2O using its molar mass (147.0146g/mol). These moles = Moles CaCl₂ = Moles of Na₂CO₃ necessaries to reach sotoichiometric quantities. To find then the mass we must use molar mass of Na₂CO₃ (105.99g/mol):
Moles CaCl₂.2H₂O:
1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂
Moles Na₂CO₃:
Mass Na₂CO₃:
0.0102 moles * (105.99g / mol) =
