# What mass of calcium when reacted with water will produce 15.7 L of hydrogen at STP?

Nov 3, 2015

$C a \left(s\right) + 2 {H}_{2} O \left(l\right) \rightarrow C a {\left(O H\right)}_{2} \left(a q\right) + {H}_{2} \left(g\right) \uparrow$

#### Explanation:

Now it is a fact that 1 mol of gas has a volume of $22.4 \cdot {\mathrm{dm}}^{3}$ at STP assuming ideal behaviour. According to the reaction above, 1 mol of calcium metal ($40.08 \cdot g$) should produce $22.4 \cdot L$ gas.

By this stoichiometry, $\frac{15.7 \cdot L \times 40.08 \cdot g \cdot m o {l}^{- 1}}{22.4 \cdot L \cdot m o {l}^{-} 1}$ $C a$ were used (about 32 g?). (NB $1 \cdot L = 1 \cdot {\mathrm{dm}}^{3}$)

Nov 3, 2015

$\text{27.7 g Ca}$ are required to produce $\text{15.7 L H"_2}$.

#### Explanation:

Balanced Equation

$\text{Ca(s)" + 2"H"_2"O}$$\rightarrow$$\text{Ca(OH)"_2("s") + "H"_2("g")}$

$\text{STP}$ is $\text{273.15 K}$ and $\text{100 kPa}$.

The molar volume of a gas at $\text{273.15 K}$ and $\text{100 kPa}$ is $\text{22.710 mol/L}$

Molar mass of $\text{Ca = 40.078 g/mol}$ (atomic weight in g/mol)

The process:

$\textcolor{red}{\text{L H"_2}}$$\rightarrow$$\textcolor{red}{\text{mol H"_2}}$$\rightarrow$$\textcolor{g r e e n}{\text{mol Ca}}$$\rightarrow$$\textcolor{b l u e}{\text{mass Ca}}$

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$\textcolor{red}{\text{L H"_2}}$$\rightarrow$$\textcolor{red}{\text{mol H"_2}}$

First determine moles $\text{H"_2}$ by dividing the given volume by the molar volume $\left(\text{22.710 L/mol}\right)$. I prefer to divide by multiplying by the reciprocal of its molar volume $\left(\text{1 mol/22.710 L}\right)$.

15.7color(red)cancel(color(black)("L H"_2))xx(1"mol H"_2)/(22.710color(red)cancel(color(black)("L")))="0.6913 mol H"_2"

$\textcolor{red}{\text{mol H"_2}}$$\rightarrow$$\textcolor{g r e e n}{\text{mol Ca}}$

To get mol $\text{Ca}$, multiply mol $\text{H"_2}$ by the mole ratio between $\text{Ca}$ and $\text{H"_2}$ from the balanced equation, with $\text{Ca}$ in the numerator.

0.6913color(red)cancel(color(black)("mol H"_2))xx(1"mol Ca")/(1color(red)cancel(color(black)("mol H")))="0.6913 mol Ca"

$\textcolor{g r e e n}{\text{mol Ca}}$$\rightarrow$$\textcolor{b l u e}{\text{mass Ca}}$

To determine mass of $\text{Ca}$ multiply times its molar mass.

0.6913color(red)cancel(color(black)("mol Ca"))xx(40.078"g Ca")/(1color(red)cancel(color(black)("mol Ca")))="27.7 g Ca" (rounded to three significant figures)

You can put all three steps together into one equation.

15.7color(red)cancel(color(black)("L H"_2))xx(1color(red)cancel(color(black)("mol H"_2)))/(22.710color(red)cancel(color(black)("L H"_2)))xx(1color(red)cancel(color(black)("mol Ca")))/(1color(red)cancel(color(black)("mol H"_2)))xx(40.078"g Ca")/(1color(red)cancel(color(black)("mol Ca")))="27.7 g Ca" (rounded to three significant figures)

Note: If your teacher is still using STP as ${0}^{\circ} \text{C}$ and $\text{1 atm}$, substitute $\text{22.414 L/mol}$ for $\text{22.710 L/mol}$.

The mass of $\text{Ca}$ would be $\text{28.1 g Ca}$.