What mass of sodium chloride, $N a C l$ $NaCl$ , is made when $200$ $200$ $L$ $L$ of chlorine gas, $C l_{2}$ $Cl_2$ , reacts with excess sodium metal?

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What mass of sodium chloride, $N a C l$ $NaCl$ , is made when $200$ $200$ $L$ $L$ of chlorine gas, $C l_{2}$ $Cl_2$ , reacts with excess sodium metal?

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Answer 1 · 2017-05-21T22:52:44

$1044$ $1044$ $g$ $g$ of $N a C l$ $NaCl$ are produced, since there are $8.93$ $8.93$ mol of $C l_{2}$ $Cl_2$ gas and each mole of $C l_{2}$ $Cl_2$ produces $2$ $2$ mol of $N a C l$ $NaCl$ .

Explanation:

One mole of an ideal gas has a volume of $22.4$ $22.4$ $L$ $L$ at STP. Chlorine is not a perfectly ideal gas but can be treated as one at this temperature and pressure.

The number of moles of gas, then, is given by $n = \frac{V}{22.4} = \frac{200}{22.4} = 8.93$ $n=V/22.4 = 200/22.4 = 8.93$ $m o l$ $mol$

Looking at the balanced equation, each $1$ $1$ mol of $C l_{2}$ $Cl_2$ yields $2$ $2$ mol of $N a C l$ $NaCl$ , so $8.93$ $8.93$ mol of $C l_{2}$ $Cl_2$ yields $2 \times 8.93 = 17.86$ $2xx8.93 =17.86$ mol of $N a C l$ $NaCl$ .

The molar mass of $N a C l$ $NaCl$ is $58.44$ $58.44$ $g$ $g$ . To find the mass of the product, we take $n = \frac{m}{M}$ $n=m/M$ and rearrange to make $m$ $m$ the product. $m = n M = 17.86 \times 58.44 = 1044$ $m=nM = 17.86xx58.44 = 1044$ $g$ $g$ (with some rounding)

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What mass of sodium chloride, $N a C l$ $NaCl$ , is made when $200$ $200$ $L$ $L$ of chlorine gas, $C l_{2}$ $Cl_2$ , reacts with excess sodium metal?

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What mass of sodium chloride, NaClNaClNaCl, is made when 200200200 LLL of chlorine gas, Cl_2Cl2Cl_2, reacts with excess sodium metal?

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Explanation:

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What mass of sodium chloride, $N a C l$ $NaCl$ , is made when $200$ $200$ $L$ $L$ of chlorine gas, $C l_{2}$ $Cl_2$ , reacts with excess sodium metal?