0.1 Analyzing chemical equations  (Page 2/3)

Problem : Determine the amount of concentrated $H_{2}S{O}_4$ required to neutralize 20 litres of ammonia gas at STP.

Solution : The chemical reaction involved is :

$$2N{H}_3+H_{2}S{O}_4\to {\left(N{H}_4\right)}_{2}S{O}_4$$

Applying mole concept :

$$\text{2 moles of}N{H}_3\equiv \text{1 mole of}{H}_{2}S{O}_4$$

$$2X22.4\text{litres of}N{H}_3\equiv 98\text{gm of}{H}_{2}S{O}_4$$

$$\text{Amount of}{H}_{2}S{O}_4\text{for 20 litres of}N{H}_3=\frac{98}{44.8}X20=43.8gm$$

Problem : Determine the amount of magnesium required to liberate 900 cc of hydrogen from a solution of HCl at 27°C and 740 mm of Hg. $A_{Mg}=24$ .

Solution : The chemical reaction involved is :

$$Mg+2HCl\to MgC{l}_2+H_2$$

Applying mole concept :

$$\text{1 mole of Mg}\equiv \text{1 mole of}{H}_2$$

$$\text{24 gm of Mg}\equiv \text{22400 cc of}{H}_2\text{at STP}$$

Using ideal gas law,

$$\Rightarrow V_2=\frac{P_1{V}_1{T}_2}{P_2{T}_1}=\frac{740X900X273}{760X300}=797cc$$

Using this data to mole relation,

$$\text{Amount of Mg}=\frac{24X797}{22400}=0.85gm$$

Problem : A 60 cc mixture of $N_{2}O$ and NO is mixed with excess $H_2$ and the resulting mixture of gas is exploded. The resulting mixture contains 38 cc of $N_2$ . Determine the volume composition of the original mixture.

Solution : The chemical reactions involved are :

$$N_{2}O+H_2\to N_2+H_{2}O$$

$$2NO+2H_2\to N_2+2H_{2}O$$

Let volume of $N_{2}O$ be x in the original mixture. Hence, volume of NO in the sample is 60 – x. Now, the corresponding mole relations are :

$$\text{x cc of}{N}_{2}O\equiv \text{x cc of}{N}_2$$

$$\text{(60−x) cc of}{N}_{2}O\equiv \frac{\left(60-x\right)}{2}\text{cc of}{N}_2$$

According to question,

$$x+\frac{60-x}{2}=38$$

$$x+60=76$$

$$\Rightarrow \text{Volume of}{N}_{2}O=x=16cc$$

$$\Rightarrow \text{Volume of NO}=60-x=60-16=44cc$$