On reaction of $38.1 \ g$ iodine and $28.4 \ g$ chlorine,a mixture of $ICl$ and $ICl_3$ is obtained. What will be the total number of moles after the reaction?

$2 SO_{2(g)} + O_{2(g)} \rightarrow 2 SO_{3(g)}$
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).

Assertion : The amount of the products formed in a reaction depends upon the limiting reagent.
Reason : Limiting reactant reacts completely in the reaction therefore limits the amount of product.

Assume carbon burns according to the following equation:
$2C_{(s)} + O_{2(g)} \rightarrow 2CO_{(g)}$
When $12 \ g$ of carbon is burnt in $48 \ g$ of oxygen,the volume of carbon monoxide produced is $...... \times 10^{-1} \ L$ at $STP$ [nearest integer].
[Given: Assume $CO$ as an ideal gas,molar mass of $C = 12 \ g \ mol^{-1}$,molar mass of $O = 16 \ g \ mol^{-1}$,and molar volume of an ideal gas at $STP$ = $22.7 \ L \ mol^{-1}$]

In the reaction $2P_2O_5 + 2HNO_3 \to P_4O_{10} + X$,what is the product $X$?

$CaCO_{3(s)} + 2 HCl_{(aq)} \rightarrow CaCl_{2(aq)} + CO_{2(g)} + H_2O_{(l)}$
Consider the above reaction,what mass of $CaCl_2$ will be formed if $250 \ mL$ of $0.76 \ M \ HCl$ reacts with $1000 \ g$ of $CaCO_3$ (in $g$)?
(Given: Molar mass of $Ca, C, O, H$ and $Cl$ are $40, 12, 16, 1$ and $35.5 \ g \ mol^{-1}$,respectively)