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saveliy_v [14]
3 years ago
13

How many moles of pcl5 can be produced from 28.0 g of p4 (and excess cl2)?

Chemistry
1 answer:
BARSIC [14]3 years ago
5 0
P₄ + 10Cl₂ ---> 4PCl₅
stoichiometry of P₄ to PCl₅ is 1:4
number of moles of P₄ reacted - 28.0 g / 124 g/mol = 0.22 mol 
Cl₂ is in excess therefore P₄ is the limiting reactant, amount of product formed depends on amount of limiting reactant present 
according to molar ratio of 1:4
number of PCl₅ moles formed  -0.22 mol x 4 = 0.88 mol 
0.88 mol of PCl₅ is formed 

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How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci
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Explanation:

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\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}.

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\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}.

Look up the ideal gas constant in the corresponding units: R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}.

Let n denote the number of moles of this gas in that V = 2\; \rm dm^{3}. By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

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Rearrange this equation and solve for n:

\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}.

In other words, there is approximately 2\; \rm mol of this gas in that V = 2\; \rm dm^{3}.

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Explanation:

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