How many moles of na2co3 are necessary to reach stoichiometric quantities? Then determine how many grams of Na2CO3 are necessary to reach stoichiometric quantities.

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) =

it is a good conductor

explanation:

elements make up everything, and they combine to make compounds.

the answer is c