Question

question 6
1.0 l of 0.35 m solutions of agno₃ and bacl₂ are mixed. the following reaction occurs:
2 agno₃(aq) + bacl₂(aq) → 2 agcl(s) + ba(no₃)₂(aq)
what is the theoretical mass of the precipitate that will be produced?
g agcl
question 7
when 15.54 g of cr is added to 0.750 l of a 0.737 m h₃po₄ solution, the following reaction occurs:
2 cr + 2 h₃po₄ → 2 crpo₄ + 3 h₂
calculate the mass of the precipitate, crpo₄, that will be produced and the moles of the reactant in excess that will remain in the solution after the reaction is complete.
g crpo₄
mol excess reactant

Explanation:

Question 6

Step1: Calculate moles of reactants

Moles of \(AgNO_3=n = c\times V\), where \(c = 0.35\space M\) and \(V=1.0\space L\). So \(n(AgNO_3)=0.35\space mol\).
Moles of \(BaCl_2=n = c\times V\), where \(c = 0.35\space M\) and \(V = 1.0\space L\). So \(n(BaCl_2)=0.35\space mol\).
From the balanced equation \(2AgNO_3(aq)+BaCl_2(aq)\to2AgCl(s)+Ba(NO_3)_2(aq)\), the mole ratio of \(AgNO_3\) to \(BaCl_2\) is \(2:1\).
For \(BaCl_2\), if \(n(BaCl_2) = 0.35\space mol\), it would require \(n(AgNO_3)=2\times0.35 = 0.7\space mol\) (but we have only \(0.35\space mol\) of \(AgNO_3\)).
For \(AgNO_3\), if \(n(AgNO_3)=0.35\space mol\), it would require \(n(BaCl_2)=\frac{0.35}{2}=0.175\space mol\) (and we have \(0.35\space mol\) of \(BaCl_2\)). So \(AgNO_3\) is the limiting reactant.

Step2: Calculate moles of \(AgCl\)

From the balanced equation, the mole ratio of \(AgNO_3\) to \(AgCl\) is \(2:2 = 1:1\). So \(n(AgCl)=n(AgNO_3)=0.35\space mol\).

Step3: Calculate mass of \(AgCl\)

The molar mass of \(AgCl\), \(M(AgCl)=107.87 + 35.45=143.32\space g/mol\).
Mass \(m=n\times M\), so \(m(AgCl)=0.35\space mol\times143.32\space g/mol = 50.162\space g\approx50.2\space g\)

Step1: Calculate moles of reactants

Moles of \(Cr\), \(n(Cr)=\frac{m}{M}\), where \(m = 15.54\space g\) and \(M(Cr)=52.00\space g/mol\). So \(n(Cr)=\frac{15.54}{52.00}=0.299\space mol\).
Moles of \(H_3PO_4=n = c\times V\), where \(c = 0.737\space M\) and \(V = 0.750\space L\). So \(n(H_3PO_4)=0.737\times0.750 = 0.553\space mol\).
From the balanced equation \(2Cr + 2H_3PO_4\to2CrPO_4+3H_2\), the mole ratio of \(Cr\) to \(H_3PO_4\) is \(1:1\).
If \(n(Cr)=0.299\space mol\), it would require \(n(H_3PO_4)=0.299\space mol\) (and we have \(0.553\space mol\) of \(H_3PO_4\)). So \(Cr\) is the limiting reactant.

Step2: Calculate moles of \(CrPO_4\)

From the balanced equation, the mole ratio of \(Cr\) to \(CrPO_4\) is \(2:2 = 1:1\). So \(n(CrPO_4)=n(Cr)=0.299\space mol\).
Molar mass of \(CrPO_4\), \(M(CrPO_4)=52.00+30.97 + 4\times16.00=156.97\space g/mol\).
Mass \(m=n\times M\), so \(m(CrPO_4)=0.299\space mol\times156.97\space g/mol\approx47.0\space g\)

Step3: Calculate moles of excess \(H_3PO_4\)

Moles of \(H_3PO_4\) used \(=n(Cr) = 0.299\space mol\)
Moles of \(H_3PO_4\) remaining \(=n_{initial}(H_3PO_4)-n_{used}(H_3PO_4)=0.553 - 0.299=0.254\space mol\)

Answer:

\(50.2\)

Question 7