Answer:
0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ are necessaries
Explanation:
<em>Based on the reaction</em>
CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃
<em>1 mole of CaCl₂ reacts per mole of Na₂CO₃</em>
<em>calculate how many moles of CaCl2•2H2O are present in 1.50 g...</em>
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):
<em>Moles CaCl₂.2H₂O:</em>
1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂
<em>Moles Na₂CO₃:</em>
<h3>0.0102 moles Na₂CO₃</h3><h3 />
<em>Mass Na₂CO₃:</em>
0.0102 moles * (105.99g / mol) =
<h3>1.08g of Na₂CO₃ are present</h3>
