What is obtained when one mole of calcium phosphide is reacted with excess water?

White phosphorus reacts with chlorine and the product hydrolyses in the presence of water. Calculate the mass of $HCl$ obtained by the hydrolysis of the product formed by the reaction of $62 \ g$ of white phosphorus with chlorine in the presence of water.

$10 \text{ g}$ of a mixture of $BaO$ and $CaO$ requires $100 \text{ cm}^3$ of $2.5 \text{ M } HCl$ to react completely. The percentage of calcium oxide in the mixture is approximately (Given: molar mass of $BaO = 153 \text{ g/mol}$,$CaO = 56 \text{ g/mol}$).

If $V \ mL$ of a substance at $NTP$ weighs $W \ g$,then the molecular weight of the substance is....

$1\, g$ of a carbonate $(M_{2}CO_{3})$ on treatment with excess $HCl$ produces $0.01186\, mol$ of $CO_{2}.$ The molar mass of $M_{2}CO_{3}$ in $g\, mol^{-1}$ is:

What is produced when $1 \ mol$ of calcium phosphide reacts with excess water?
$(Hint: Ca_3P_2 + H_2O (excess) \rightarrow Ca(OH)_2 + 2PH_3)$