Chemistry Test - 3

A sample of calamine ore \(\left(Z n C O_{3}\right)\) contains clay as impurity which is \(37.5 \%.\)
Let the total mass of the sample be \(100 g\)
\(\therefore m_{Z n C O_{3}}=62.5 g\) and
\(m_{c l a y}=37.5 g\)
\(Z n C O_{3} \rightarrow Z n O+C O_{2.}\)
\(125 g\) of \(ZnCO _{3}\) corresponds to \(44 g\) of \(CO _{2.}\)
\(\therefore 62.5 g\) corresponds to \(22 g\) of \(C O_{2.}\)
\(m_{H_{2} O}+m_{C O_{2}}=28 g\)
\(\therefore m_{H_{2} O}=28-22=6 g\)
\(\therefore \%\) loss in the weight of clay \(=\frac{6}{37.5} \times 100=16 \%\)

When silver carbonate is strongly heated, a residue of silver is obtained.
The reaction is as follows:
\(A g_{2} C O_{3}(s) \stackrel{\text {Heat}}{\longrightarrow} 2 A g(s)+C O_{2}(g)+\frac{1}{2} O_{2}(g)\)
The molecular weight of silver carbonate and silver is \(276 g / mol\) and \(108 g / mol\) respectively. Thus, when \(276 g\) (or \(1 mole\) ) of silver carbonate is heated, \(2 \times 108=216 g (2 moles )\) of silver is obtained. When \(2.76 g\) of silver carbonate is heated, the amount of residue (silver) obtained is \(\frac{216}{276} \times 2.76=2.16 g.\)

o.i \(M NaCl\)
\(N a C l \rightarrow N a^{+}+C l^{-}\)
I mol of \(N a C l=1\) mol of \(C l\)
In 1 ltr solution, o.t mol of \(N a C l\)
\(=0.1 mol\) of \(C l^{-}\)
\(0.2 M CaCl _{2}-2 ltr\)
\(CaCl _{2} \rightarrow Ca ^{+}+2 Cl^{-}\)
I mol of \(C a C l_{2}=2\) mol of \(C l^{-}\)
In 1 ler solution, 0.2 mol of \(C a C l_{2}=0.2 \times 2 mol\) of \(C l^{-}\)
In 2 ltr solution \(=0.8\) mol of \(C l^{-}\)
Total moles of \(C l^{-}\)
\(=00.8+0.1=0.9 mol\)
Desired Molarity of \(C l^{-}=0.34 M\)
So, volume of solution \(=\) \(\frac{0.9 mol }{0.34 M }=2.65\) litr

In the first oxide, the percentages of \(O\) and \(S\) are \(50 \%\) and \(50 \%\) respectively.
Thus, \(1 g\) of oxygen will combine with \(1 g\) of sulphur.
In the second oxide, the percentages of \(O\) and \(S\) are \(60 \%\) and \(40 \%\), respectively.
Thus, \(1 g\) of oxygen will combine with \(\frac{2}{3} g\) of sulphur.
The ratio of weights of sulphur which combine with 1 g of oxygen \(=1: \frac{2}{3}=3: 2\)

\(N a_{2} S_{2} O_{3} \rightarrow N a_{2} S O_{4}\)
The total change in oxidation number \(=4 \times 2=8\) \(\therefore E_{N a_{2} S_{2} O_{3}}=\frac{m o l . w t}{V . f}=\frac{M}{8}\)

The balanced reaction is
\[CaCO _{3}(s)+2 HCl (a q) \rightarrow CaCl _{2}(a q)+C O_{2}(g)+H_{2} O(l)
\]\(10 g\) calcium carbonate corresponds to \(\frac{10}{100}=0.1 mole\) (Molecular weight of \(CaCO _{3}=100 g / mol)\).
It will react with \(0.2 mol HCl\).
However, \(0.3 mol HCl\) is present.
Hence, calcium carbonate is the limiting reagent.
\(0.1 mol\) of calcium carbonate will form \(0.1 mol\) of calcium chloride.

The solution is \(0.15 N\). Hydrogen chloride (HCl) contains one replaceable H atom.
Molar mass of \(HCl\)\(=(1+35.5)=36.5 u\)
The expression for the equivalent weight of an acid is as given below.
Equivalent weight=Molecular mass/Number of replaceable H
HCl contains one replacable hydrogen. Substitute values in the above expression.
\(=\frac{36.5}{1}=36.5 u\)
The expression for the normality of solution is as given below.
=Solute in grams per litre of solution/Equivalent weight in grams
Solute in grams per litre of solution = Normality Equivalent weight in grams.
The mass of pure hydrogen chloride gas dissolved per litre of its solution
\(=(0.15 \times 36.5) g=5.475 g / l\)

The ratio of the volumes of dinitrogen, oxygen and nitrous oxide is \(45.4 : 22.7 : 45.4.\)
This is in the simple whole number ratio 2 : 1 : 2
This satisfies the Gay Lussac's law of combining volume of gases.

The balanced reaction is given below:
\(2 S O_{2}+O_{2}+2 H_{2} O \rightarrow 2 H_{2} S O_{4}\)
\(\begin{array}{ll}\text { 5.6 } 4.8 & \text { (No. of moles given) }\end{array}\)
\(\frac{5.6}{2} \quad \frac{4.8}{1} \quad\) (Ratio of moles given and stoichiometry coefficient) \(=2.8=4.8\)
As 2.8 is lower than \(4.8, S O_{2}\) is the limiting reagent and product will form according to number of moles of \(S O_{2}\) (given).
So, moles produced of \(H_{2} S O_{4}=\frac{2}{2} \times 5.6=5.6 mol\)

Molecular formula of calcium
phosphate is \(C a_{3}\left(P O_{4}\right)_{2}\)
Its molar mass is \(3(40)+2(31)+8(16)=310 g / m o l\)
Percent of calcium \(=\frac{120}{310} \times 100=38.71 \%\)
Percent of phosphorus \(=\frac{62}{310} \times 100=20.0 \%\)
Percent of oxygen \(=\frac{128}{310} \times 100=41.29 \%\)