Since we are already given the balanced equation:
→
We can derive the molar ratios as: 1:2:1:1
That being said, we are given 0.172 moles of bromine (), so it has a ratio of 1:1 with sodium bromide (
).
So we can take from that ratio, that when 0.172 moles of bromine are used, we are, in turn, going to get 0.172 moles of sodium bromide produced.
Answer:
Explanation:
Hello,
At first, we compute liquid-phase molar fractions:
Now, by means of the fugacity concept, for hexane, for instance, we have:
In this manner, at 25 °C the vapor pressure of hexane and heptane are 0.198946 atm and 0.013912 atm repectively, thus, the total pressure is:
Finally, from the hexane's fugacity equation, we find its mole fraction in the vapour as:
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Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Thus, we proceed to compute the reacting moles of Pb(NO3)2 and KI, by using the given concentrations and densities and molar masses which are 331.2 g/mol and 166 g/mol respectively:
Next, the 0.0464 moles of Pb(NO3)2 will consume the following moles of KI (consider their 1:2 molar ratio):
Hence, as only 0.0789 moles of KI are available, KI is the limiting reactant, therefore the formed grams of PbI2, considering its molar mass of 461.01 g/mol and 2:1 molar ratio, are:
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