The answer is A because it’s how you calculate the mass
Answer:
74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.
Explanation:
The balanced reaction is:
Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:
- Na₂CO₃: 1 mole
- Ca(NO₃)₂: 1 mole
- CaCO₃: 1 mole
- NaNO₃: 2 mole
Being the molar mass of the compounds:
- Na₂CO₃: 106 g/mole
- Ca(NO₃)₂: 164 g/mole
- CaCO₃: 100 g/mole
- NaNO₃: 85 g/mole
then by stoichiometry the following quantities of mass participate in the reaction:
- Na₂CO₃: 1 mole* 106 g/mole= 106 g
- Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
- CaCO₃: 1 mole* 100 g/mole= 100 g
- NaNO₃: 2 mole* 85 g/mole= 170 g
You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of CaCO₃?

mass of CaCO₃= 74.81 grams
<u><em>74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.</em></u>
0.347 mols, working out shown on photo
Glycerol attractive forces are great than water. The harder to break, the more energy is needed.
K₃PO₄ → 3K⁺ (aq) + PO₄³⁻(aq)
One mole of PO₄³⁻ ion gets dissociated from one mole of K₃PO₄
As per the definition of Avogadro's number, 1 mole = 6.022 x 10²³ ions
One mole of PO₄³⁻ ions x (6.022 x 10²³ ions/ 1 mole of PO₄³⁻ ions )
= 6.022 x 10²³ ions
Therefore , there are 6.022 x 10²³ PO₄³⁻ ions in a mole of K₃PO₄.