I need help with determing the mass of a reaction containing Cu and Cl2 with iron nails Given the information: mass of the empty dry beaker: 105.52 , initial mass of the dry beaker +CuCl2(s) =...
I need help with determing the mass of a reaction containing Cu and Cl2 with iron nails
Given the information: mass of the empty dry beaker: 105.52 , initial mass of the dry beaker +CuCl2(s) = 113.02g, Initial mass of the 2 iron nails = 3.82g, Final mass of the dry beaker+Cu= 106.41g, Final mass of 2 iron nails=3.04gWhat mass of CuCl2 (s) is used?
What mass of iron is used in the reaction?
How many moles of iron are used?
What mass of Cu is produced in the reaction?
How many moles of Cu are produced?
Find the ratio of the moles of copper produced to moles of iron used.
Determine the number of atoms of iron reacting and the number of atoms of copper produced.
The blue colour of the solution gets lighter as the reaction proceeds. What does this indicate? Explain your answer.
Write a balanced equation for the reaction. This is a single replacement reaction.
How many moles, how many atoms, and what mass of Cu (s) are produced by the reaction when 100. grams of iron completely reacts, assuming there is enough CuCl2 (aq)?
CuCl2 + Fe --> FeCl2 + Cu + Fe
For every mole of Fe used, one mole of CuCl2 is also used.
1. Since from #3 there were 0.014 moles of Fe used, then there were 0.014 moles of CuCl2 * 134.45 g/mol = 1.88 g of CuCl2 used.
2. mass of iron. 3.82 g - 3.04 g = .78 g of Fe used.
3. Moles of iron used: .78/55.85 = 0.014 moles of Fe
4. Mass of Cu produced: 106.41-105.52 = 0.89 g of Cu
5. moles of Cu: 0.89/63.55 = 0.014 moles of Cu
6. mole ratio: 0.014: 0.014 = 1:1 ratio as expected from the balanced chemical equation.
7.# of atoms of Fe &Cu. 0.014 moles * 6.023x10^23 atoms/mole = 8.43 x 10^21 atoms of each.
8. Blue color fades because the CuCl2 concentration is decreasing.
9. Balanced equation is at top.
10. moles, mass, atoms if use 100 g of Fe. 100/55.85 = 1.79 moles of Fe used. So produce 1.79 moles of Cu, which is 1.79*63.55 = 113.75 g of Cu. 1.79 * 6.023x 10^23 = 1.078 x 10^24 atoms of Cu.