Answer: d. More than 6.5 grams of copper (II) is formed, and some copper chloride is left in the reaction mixture.
Explanation: 
As can be seen from the chemical equation, 2 moles of aluminium react with 3 moles of copper chloride.
According to mole concept, 1 mole of every substance weighs equal to its molar mass.
Aluminium is the limiting reagent as it limits the formation of product and copper chloride is the excess reagent as (14-7.5)=6.5 g is left as such.
Thus 54 g of of aluminium react with 270 g of copper chloride.
1.50 g of aluminium react with=
of copper chloride.
3 moles of copper chloride gives 3 moles of copper.
7.5 g of copper chloride gives 7.5 g of copper.
<span>Answer:
For this problem, you would need to know the specific heat of water, that is, the amount of energy required to raise the temperature of 1 g of water by 1 degree C. The formula is q = c X m X delta T, where q is the specific heat of water, m is the mass and delta T is the change in temperature. If we look up the specific heat of water, we find it is 4.184 J/(g X degree C). The temperature of the water went up 20 degrees.
4.184 x 713 x 20.0 = 59700 J to 3 significant digits, or 59.7 kJ.
Now, that is the energy to form B2O3 from 1 gram of boron. If we want kJ/mole, we need to do a little more work.
To find the number of moles of Boron contained in 1 gram, we need to know the gram atomic mass of Boron, which is 10.811. Dividing 1 gram of boron by 10.811 gives us .0925 moles of boron. Since it takes 2 moles of boron to make 1 mole B2O3, we would divide the number of moles of boron by two to get the number of moles of B2O3.
.0925/2 = .0462 moles...so you would divide the energy in KJ by the number of moles to get KJ/mole. 59.7/.0462 = 1290 KJ/mole.</span>
Answer:
The correct answer is is option B
b. 93.3 g
Explanation:
SEE COMPLETE QUESTION BELOW
Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s)
How many grams of HCl can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl?
a. 7.30 g
b. 93.3 g
c. 146 g
d. 150 g
e. 196 g
CHECK THE ATTACHMENT FOR STEP BY STEP EXPLANATION
Electrochemical cell representation for above reaction is,
Br-/Br2//I2/I-
Reaction at Anode: Br2 + 2e- → 2Br- (1)
Reaction at Cathode: 2I- → I2 + 2e- (2)
Standard reduction potential for Reaction 1 = Ered(anode) = 1.066 v
Standard reduction potential for Reaction 2 = Ered(cathode) = 0.535 v
Eo cell = Ered(cathode) - Ered(anode)
= 0.535 - 1.066
= -0.531v
Now, we know that ΔGo = -nF (Eo cell) ..............(3)
Also, ΔGo = RTln(K) ..........(4)
Equation 3 and 4 we get,
ln (K) = nF (Eo cell) / RT
= 2 X 96500 X (-0.531)/ (8.314 X 298)
∴ K = 1.085 X 10^-18.
