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Question

$$120$$g Mg is burnt in air to give a mixture of $$MgO$$ and $$Mg_3N_2$$. The mixture is now dissolved in $$HCl$$ to form $$MgCl_2$$ and $$NH_4Cl$$. If $$107$$g $$NH_4Cl$$ is produced, then determine the moles of $$MgCl_2$$ formed.


Solution

Number of moles of $$NH_4Cl=\cfrac {107 gm}{53.4 g mol^{-1}}=2 mol$$
$$2 Mg+O_2\rightarrow 2MgO$$      $$3Mg+N_2 \rightarrow Mg_3N_2$$
$$MgO+2HCl\rightarrow MgCl_2+H_2O$$
$$MgO+2HCl \rightarrow MgCl_2+H_2O$$
$$Mg_3N_2+8HCl\rightarrow 3MgCl_2+2NH_4Cl$$
$$120g Mg=\cfrac {120}{24}=5$$ moles of $$Mg$$
So $$2$$ moles of $$Mg$$ gives $$2$$ moles of $$MgO$$ and $$3$$ moles of $$Mg$$ gives one mole of $$Mg_3N_2$$. $$1$$ mole $$MgO$$ and $$1$$ mole $$Mg_3N_2$$ give one mole and $$3$$ mole $$MgCl_2$$ respectively. Again one mole $$Mg_3N_2$$ give $$2$$ moles of $$NH_4Cl$$ .
The moles of $$MgCl_2$$ formed= $$2 \times 1+3=5$$ moles

Chemistry

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