Question

5.3 gram of pure NaCO3 reacts with 7 gram of HCL to produce NACL ,H2O and CO2
find the limiting reagent.
1.calculate the no. of modes of excess reactant left over unreacts.
2.what volume of CO2 formed at NTP.
3.Find the no.of molecules of H2O formed.​

Answer 1

first we need to write the balanced chemical equation

Na2CO 3 +2 HCl -->2NaCl +H 2 O + C O 2

mass of Na 2 CO 3 = 5.3 gram

molar mass of Na 2 CO 3 =106 g

moles of Na 2 CO 3

= mass / molar mass

= 5.3 / 106

= 0.05

moles of Na 2 CO 3=0.05

mass of HCL =7 g

molar mass of HCl = 36.5 g

moles of HCl = 7 / 36.5 =0.2

moles of HCl = 0.2

As per the above balanced equation we can conclude that ,

If 1 mole of Na 2 CO 3 reacts with 2 moles of HCl then

0.05 moles of Na 2 CO 3 will react with 0.05×2=0.1 mole of HCl

available amount of HCl is 0.2 hence HCl is not a limiting reagent

Similarly ,

If 2 mole of HCl reacts with 1 moles of Na 2 CO 3 then

0.2moles of HCl will react with 0.2/2=0 .1 mole of Na 2 CO 3

available amount of Na 2 CO 3 is 0.05 hence

Na 2 CO 3 is a limiting reagent

If 1 mole of Na 2 CO 3 gives 2 moles of NaCl

then 0.05 moles of Na 2 CO 3 will give 0.1 moles of NaCl

moles of NaCl=0.1

Similarly ,

If 1 mole of Na 2 CO 3 gives 1 mole of H 2O

then 0.05 moles of Na 2 CO 3 will give 0.05 moles of H 2 O

moles of H 2O = moles of CO 2= 0.05

moles of CO 2 = V CO2 / 22.4 L

V CO 2 = 0.05 × 22.4

V CO 2 =1.12 L

moles of H 2 O = molecule of H 2 O / N a

molecule of H 2 O = 0.05 × 6.023×10^23

molecule of H 2 O= 3.011×10^22