Balance the following chemical equations:
$(i)$ $Cu_2S + O_2 \to Cu_2O + SO_2$
$(ii)$ $Ca_3(PO_4)_2 + H_2SO_4 \to Ca(H_2PO_4)_2 + CaSO_4$

For the reaction,$2Fe(NO_3)_3 + 3Na_2CO_3 \to Fe_2(CO_3)_3 + 6NaNO_3$. Initially,if $2.5 \ mol$ of $Fe(NO_3)_3$ and $3.6 \ mol$ of $Na_2CO_3$ are taken. If $6.3 \ mol$ of $NaNO_3$ is obtained,then the $\%$ yield of the given reaction is:

The mass of $N_2F_4$ produced by the reaction of $2.0 \ mol$ of $NH_3$ and $8.0 \ mol$ of $F_2$ is $0.5 \ mol.$ What is the per cent yield?
$2NH_3 + 5F_2 \to N_2F_4 + 6HF$

Equal weights of $X$ (atomic weight $= 36$) and $Y$ (atomic weight $= 24$) are reacted to form the compound $X_2Y_3$. Which of the following is correct?

$92 \ g$ of $NO_{2(g)}$ and $32 \ g$ of $O_{2(g)}$ are taken in a rigid container. At a constant temperature,they react to produce $N_2O_{5(g)}$. The pressure in the container after the reaction is observed to be $\frac{7}{8} P$,where $P$ is the initial pressure of the gaseous mixture. The percentage yield of the reaction is ....$\%$

$2 SO_{2(g)} + O_{2(g)} \rightarrow 2 SO_{3(g)}$
The above reaction is carried out in a vessel starting with partial pressures $P_{SO_{2}} = 250 \ mbar$,$P_{O_{2}} = 750 \ mbar$,and $P_{SO_{3}} = 0 \ mbar$. When the reaction is complete,the total pressure in the reaction vessel is $..... \ mbar$. (Round off to the nearest integer).