Answer:
#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.
Explanation:
Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆
1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.
=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP
2. Calculate moles of F₂ used
=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used
3. Calculate moles of XeF₆ produced from reaction ratios …
Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆
4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number (6.02 x 10²³ molecules/mole)
=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)
= 2.75 x 10²³ molecules XeF₆.
The balanced chemical reaction would be as follows:
<span>5P4O6 +8I2 ---> 4P2I4 +3P4O10
We are given the amount of reactants used for the reaction. We first need to determine the limiting reactant from the given amounts. We do as follows:
8.80 g P4O6 (1 mol / </span><span>219.88 g) = 0.04 mol P4O6
12.37 g I2 ( 1 mol / </span><span>253.809 g ) = 0.05 mol I2
Therefore, the limiting reactant is iodine since less it is being consumed completely in the reaction. We calculate the amount of P2I4 prepared as follows:
0.05 mol I2 ( 4 mol P2I4 / 8 mol I2 ) (</span><span>569.57 g / 1 mol) = 14.24 g P2I4</span>
Answer:
Percentage mass of copper in the sample = 32%
Explanation:
Equation of the reaction producing Cu(NO₃) is given below:
Cu(s)+ 4HNO₃(aq) ---> Cu(NO₃)(aq) + 2NO₂(g) + 2H₂O(l)
From the equation of reaction, 1 mole of Cu(NO₃) is produced from 1 mole of copper. Therefore, 0.010 moles of Cu(NO₃) will be produced from 0.010 mole of copper.
Molar mass of copper = 64 g/mol
mass of copper = number of moles * molar mass
mass of copper = 0.01 mol * 64 g/mol = 0.64 g
Percentage by mass of copper in the 2.00 g sample = (0.64/2.00) * 100%
Percentage mass of copper in the sample = 32%
