Question Number 191897 by sciencestudentW last updated on 03/May/23
$${if}\:{we}\:{combine}\:\mathrm{100}{gr}\:{Na}\:{with}\:\mathrm{180}\:{Cl}_{\mathrm{2}} \\ $$$${then}\:{how}\:{much}\:{NaCl}\:{will}\:{produce}? \\ $$
Answered by Subhi last updated on 03/May/23
$$\mathrm{252}.\mathrm{17}\:{g} \\ $$$$ \\ $$
Commented by sciencestudentW last updated on 03/May/23
$${practice}? \\ $$
Commented by sciencestudentW last updated on 03/May/23
$${please}\:{show}\:{your}\:{practice}! \\ $$
Answered by Subhi last updated on 03/May/23
$${no}\:{of}\:{moles}\:{of}\:{Na}=\:\frac{\mathrm{100}}{\mathrm{23}} \\ $$$${moles}\:{of}\:{Cl}_{\mathrm{2}} =\:\frac{\mathrm{180}}{\mathrm{70}} \\ $$$$\mathrm{2}{Na}+{Cl}_{\mathrm{2}} \Rightarrow\mathrm{2}{NaCl} \\ $$$${Na}:{NaCl}\:\Rightarrow\mathrm{2}:\mathrm{2}\Rightarrow\frac{\mathrm{100}}{\mathrm{23}}:\frac{\mathrm{100}}{\mathrm{23}} \\ $$$${Cl}_{\mathrm{2}} :{Nacl}\Rightarrow\mathrm{1}:\mathrm{2}\Rightarrow\frac{\mathrm{18}}{\mathrm{7}}:\frac{\mathrm{36}}{\mathrm{7}} \\ $$$${Na}\:{is}\:{the}\:{limiting}\:{reactant}\:{in}\:{the}\:{reaction} \\ $$$${no}\:{of}\:{moles}\:{of}\:{NaCl}=\mathrm{100}/\mathrm{23} \\ $$$${mass}={moles}×{molar}\:{mass}=\frac{\mathrm{100}}{\mathrm{23}}×\left(\mathrm{35}+\mathrm{23}\right)=\mathrm{252}.\mathrm{17}\:{g} \\ $$$${remained}\:{of}\:{Cl}_{\mathrm{2}} =\frac{\mathrm{18}}{\mathrm{7}}−\left(\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{100}}{\mathrm{23}}\right)=\mathrm{0}.\mathrm{39} \\ $$$${mass}\:{remaied}=\mathrm{0}.\mathrm{39}×\mathrm{70}=\mathrm{27}.\mathrm{826}\:{g} \\ $$