# Question f72ba

May 3, 2015

Your reaction will produce 0.624 g of hydrogen gas.

Start by writing the balanced chemical equation for this reaction

$C {a}_{\left(s\right)} + 2 {H}_{2} {O}_{\left(l\right)} \to C a {\left(O H\right)}_{\textrm{2 \left(s\right]}} + {H}_{\textrm{2 \left(g\right]}} \uparrow$

Notice that you have a $1 : 1$ mole ratio between calcium and hydrogen gas. This tells you that, regardless of how many moles of calcium react, you produce an equal number of moles of hydrogen gas.

Determine how many moles of calcium reacted by using its molar mass

12.4cancel("g") * "1 mole Ca"/(40.078cancel("g")) = "0.3094 moles Ca"

Because water is in excess, i.e. it will not act as a limiting reagent, you can assume that all of the calcium will react. This means that your reaction will produce

0.3094cancel("moles Ca") * ("1 mole "H_2)/(1cancel("mole Ca")) = "0.3094 moles " ${H}_{2}$

Now use hydrogen gas' molar mass to see how many grams you'd get

0.3094cancel("moles "H_2) * ("2.016 g "H_2)/(1cancel("mole "H_2)) = "0.6238 moles "# ${H}_{2}$

Rounded to three sig figs, the answer will be

${m}_{{H}_{2}} = \textcolor{g r e e n}{\text{0.624 g}}$