Respuesta :
Answer:
99.56 g.
Explanation:
- The balanced reaction is:
3Ca(NO₃)₂ + 2Na₃PO₄ → 6NaNO₃ + Ca₃(PO₄)₂.
It is clear that 3.0 moles of Ca(NO₃)₂ react with 2.0 moles of Na₃PO₄ to produce 6.0 moles of NaNO₃ and 1.0 mole of Ca₃(PO₄)₂.
- We need to calculate the no. of moles of 96.1 g of Ca(NO₃)₂:
n = mass/molar mass = (96.1 g)/(164.088 g/mol) = 0.5857 mol.
Using cross multiplication:
3.0 moles of Ca(NO₃)₂ produce → 6.0 moles of NaNO₃.
0.5857 moles of Ca(NO₃)₂ produce → ??? moles of NaNO₃.
∴ The no. of moles of NaNO₃ = (6.0 mol)(0.5857 mol)/(3.0 moles) = 1.171 mol.
∴ The no. of grams of NaNO₃ = (no. of moles of NaNO₃)(molar mass) = (1.171 mol)(84.99 g/mol) = 99.56 g.
Answer: The mass of calcium phosphate will be, 61.07 grams
Explanation : Given,
Mass of calcium nitrate = 96.1 g
Molar mass of calcium nitrate = 164 g/mole
Molar mass of calcium phosphate = 310 g/mole
First we have to calculate the moles of calcium nitrate.
[tex]\text{Moles of }Ca(NO_3)_2=\frac{\text{Mass of }Ca(NO_3)_2}{\text{Molar mass of }Ca(NO_3)_2}=\frac{96.1g}{164g/mole}=0.59moles[/tex]
Now we have to calculate the moles of calcium phosphate.
The balanced chemical reaction will be,
[tex]3Ca(NO_3)_2+2Na_3PO_4\rightarrow 6NaNO_3+Ca_3(PO_4)_2[/tex]
From the balanced chemical reaction, we conclude that
As, 3 moles of calcium nitrate react to give 1 mole of calcium phosphate
So, 0.59 moles of calcium nitrate react to give [tex]\frac{1}{3}\times 0.59=0.197[/tex] moles of calcium phosphate
Now we have to calculate the mass of calcium phosphate.
[tex]\text{Mass of }Ca_3(PO_4)_2=\text{Moles of }Ca_3(PO_4)_2\times \text{Molar mass of }Ca_3(PO_4)_2[/tex]
[tex]\text{Mass of }Ca_3(PO_4)_2=0.197moles\times 310g/mole=61.07g[/tex]
The theoretical yield of calcium phosphate = 61.07 g
Therefore, the mass of calcium phosphate will be, 61.07 grams
