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

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

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Answer 1 · 2015-11-03T21:22:17

$C a (s) + 2 H_{2} O (l) \to C a {(O H)}_{2} (a q) + H_{2} (g) ↑ ⏐$ $Ca(s) + 2H_2O(l) rarr Ca(OH)_2(aq) + H_2(g)uarr$

Explanation:

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

By this stoichiometry, $\frac{15.7 \cdot L \times 40.08 \cdot g \cdot m o l^{- 1}}{22.4 \cdot L \cdot m o l^{- 1}}$ $(15.7*Lxx40.08*g *mol^(-1))/(22.4*L*mol^-1)$ $C a$ $Ca$ were used (about 32 g?). (NB $1 \cdot L = 1 \cdot d m^{3}$ $1*L=1*dm^3$ )

Answer 2 · 2015-11-03T22:48:42

$27.7 g Ca$ $"27.7 g Ca"$ are required to produce ${15.7 L H}_{2}$ $"15.7 L H"_2"$ .

Explanation:

Balanced Equation

$Ca(s) + 2 H_{2} O$ $"Ca(s)" + 2"H"_2"O"$ $\to$ $rarr$ ${Ca(OH)}_{2} (s) + H_{2} (g)$ $"Ca(OH)"_2("s") + "H"_2("g")"$

$STP$ $"STP"$ is $273.15 K$ $"273.15 K"$ and $100 kPa$ $"100 kPa"$ .

The molar volume of a gas at $273.15 K$ $"273.15 K"$ and $100 kPa$ $"100 kPa"$ is $22.710 mol/L$ $"22.710 mol/L"$

Molar mass of $Ca = 40.078 g/mol$ $"Ca = 40.078 g/mol"$ (atomic weight in g/mol)

The process:

${L H}_{2}$ $color(red)("L H"_2")$ $\to$ $rarr$ ${mol H}_{2}$ $color(red)("mol H"_2")$ $\to$ $rarr$ $mol Ca$ $color(green)("mol Ca")$ $\to$ $rarr$ $mass Ca$ $color(blue)("mass Ca")$

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${L H}_{2}$ $color(red)("L H"_2")$ $\to$ $rarr$ ${mol H}_{2}$ $color(red)("mol H"_2")$

First determine moles $H_{2}$ $"H"_2"$ by dividing the given volume by the molar volume $(22.710 L/mol)$ $("22.710 L/mol")$ . I prefer to divide by multiplying by the reciprocal of its molar volume $(1 mol/22.710 L)$ $("1 mol/22.710 L")$ .

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

$color(red)("mol H"_2")$ $rarr$ $color(green)("mol Ca")$

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

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

$color(green)("mol Ca")$ $rarr$ $color(blue)("mass Ca")$

To determine mass of $"Ca"$ 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 $0^@"C"$ and $"1 atm"$ , substitute $"22.414 L/mol"$ for $"22.710 L/mol"$ .

The mass of $"Ca"$ would be $"28.1 g Ca"$ .

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

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