Answer:
24.5 g of NaCl
Explanation:
We begin from the balanced reaction:
3MgCl₂ + 2Na₃PO₄ → 6NaCl + Mg₃(PO₄)₂
If the sodium phosphate is in excess, then the limting reagent is the magnessium chloride.
We convert mass to moles:
20 g . 1mol / 95.2g = 0.210 moles.
3 moles of MgCl₂ can produce 6 moles of NaCl
0.210 moles of salt, may produce (0.210 . 6) /3 = 0.420 moles
Ratio of reactant is twice the product
We convert the moles to mass:
0.420 mol . 58.45 g/mol = 24.5 g
Nitrate
Hope this helps :))
Some
of the solutions exhibit
colligative properties. These properties depend on the amount of solute
dissolved in a solvent. These properties include freezing point depression, boiling
point elevation, osmotic pressure and vapor pressure lowering. Calculations
are as follows:
<span>
ΔT(freezing point) = (Kf)mi
3 = 1.86 °C kg / mol (m)(2)
3 =3.72m
m = 0.81 mol/kg</span>
