Boyle’s Law P1V1 = P2V2
P1 = 0.80 atm V1 = 1.8 L
P2 = 1.0 atm V2 = ??
(.8 atm)(1.8 L) = (1.0 atm)(V2)
1.44 atm x L = 1 atm V2
Answer:
Limiting reactant: Cl2.
Max mass of PCl3 = 41.44 g.
P4 leftover = 5.51 g.
Explanation:
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In this case, since the undergoing chemical reaction is:
In order to compute the maximum amount of PCl3, it is necessary to compute the grams of this product produced by each reactant, just as shown below, whereas molar masses and mole ratios are used:
Thus, since chlorine gas yields fewer grams of PCl3 than P4 we infer Cl2 is the limiting reactant and 63.41 grams of PCl3 product are yielded.
Finally, for the excess reactant, we see a difference of 63.41-41.44=21.97g, so we can compute of the leftover of P4 as follows:
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10.53 grams is the theoretical yield of CaF2 for this experiment.
Explanation:
From the reaction it can be found that Ca reacts with excess of fluoride gas to form calcium fluoride. Thus, it can be concluded that calcium is limiting reagent as product formation depends on the concentration of calcium in the reaction as reactant.
The balanced chemical reaction is :
Ca + F2 ⇒ CaF2
Thus one mole of calcium reacted to form one mole of calcium fluoride.
Number of moles of calcium fluoride can be obtained by using the formula;
number of moles=
=
= 0.135 moles of CaF2 is formed.
So,
1 mole of calcium reacts to give 1 mole of CaF2
x moles of calcium will give 0.135 moles of CaF2
Hence, 0.135 moles of CaF2 is formed
the weight formed will be given as 0.135 x 78.07
= 10.53 grams
The theoretical yield is the amount of product formed by complete conversion of limiting reactant. Thus, 10.53 grams is the theoretical yield of CaF2 in the reaction.
When electricity runs through an electric stove, what other type of energy is produced?
Answer: A.) Heat Energy