Respuesta :
Answer:- 60.6 g
Solution:- The balanced equation for the given reaction is:
[tex]2Na_3PO_4(aq)+3Ca(NO_3)_2(aq)\rightarrow 6NaNO_3(aq)+Ca_3(PO_4)_2(s)[/tex]
It is a grams to grams conversion problem. Here, first of all the given grams of calcium nitrate are converted to moles on diving by its molar mass. Then it is multiplied by the mol ratio to get the moles of calcium phosphate and finally multiplied by molar mass of calcium phosphate to get its grams.
From balanced equation, there is 3:1 mol ratio between calcium nitrate and calcium phosphate.
Molar mass of calcium nitrate is 164.088 g per mol and the molar mass of calcium phosphate is 310.18 g per mol.
[tex]96.1gCa(NO_3)_2(\frac{1molCa(NO_3)_2}{164.088gCa(NO_3)_2})(\frac{1molCa_3(PO_4)_2}{3molCa(NO_3)_2})(\frac{310.18gCa_3(PO_4)_2}{1molCa_3(PO_4)_2})[/tex]
= [tex]60.6gCa_3(PO_4)_2[/tex]
So, the mass of calcium phosphate the reaction can produce is 60.6 g.
Answer is: The mass of calcium phosphate the reaction can produce is 60.8 grams.
Balanced chemical reaction:
3Ca(NO₃)₂(aq) + 2Na₃PO₄(aq) → 6NaNO₃(aq) + Ca₃(PO₄)₂(s).
m(Ca(NO₃)₂) = 96.1 g; mass of calcium nitrate.
n(Ca(NO₃)₂) = m(Ca(NO₃)₂) ÷ M(Ca(NO₃)₂).
n(Ca(NO₃)₂) = 96.1 g ÷ 164.1 g/mol.
n(Ca(NO₃)₂) = 0.58 mol; amount of calcium nitrate.
From chemical reaction: n(Ca(NO₃)₂) : nCa₃(PO₄)₂ = 3 : 1.
n(Ca₃(PO₄)₂) = 0.58 mol ÷ 3.
n(Ca₃(PO₄)₂) = 0.196 mol; amount of calcium phosphate.
m(Ca₃(PO₄)₂) = n(Ca₃(PO₄)₂) · M(Ca₃(PO₄)₂).
m(Ca₃(PO₄)₂) = 0.196 mol · 310.18 g/mol.
m(Ca₃(PO₄)₂) = 60.8 g; mass of calcium phosphate.
