Answer:
Option D. 0.115 M
Explanation:
The following data were obtained from the question:
Mass of CuSO4 = 36.8 g
Volume of solution = 2 L
Molar mass of CuSO4 = 159.62 g/mol
Molarity of CuSO4 =..?
Next, we shall determine the number of mole in 36.8 g of CuSO4.
This can be obtained as shown below:
Mass of CuSO4 = 36.8 g
Molar mass of CuSO4 = 159.62 g/mol
Mole of CuSO4 =.?
Mole = mass /Molar mass
Mole of CuSO4 = 36.8 / 159.62
Mole of CuSO4 = 0.23 mole
Finally, we shall determine the molarity of the CuSO4 solution as illustrated below:
Mole of CuSO4 = 0.23 mole
Volume of solution = 2 L
Molarity of CuSO4 =..?
Molarity = mole /Volume
Molarity of CuSO4 = 0.23 / 2
Molarity of CuSO4 = 0.115 M
Therefore, the molarity of the CuSO4 solution is 0.115 M.
1 molecule of NO2 included N atom and 2 oxygen atoms, so a total of 3 atoms per molecule.
1.3 x 10^22 times 3 = 3.9 x 10^22 atoms
Answer:
a. NH3 is limiting reactant.
b. 44g of NO
c. 40g of H2O
Explanation:
Based on the reaction:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l)
4 moles of ammonia reacts with 5 moles of oxygen to produces 4 moles of NO and 6 moles of water.
To find limiting reactant we need to find the moles of each reactant and using the balanced equation find which reactant will be ended first. Then, with limiting reactant we can find the moles of each reactant and its mass:
<em>a. </em><em>Moles NH3 -Molar mass. 17.031g/mol-</em>
25g NH3*(1mol/17.031g) = 1.47moles NH3
Moles O2 = 4 moles
For a complete reaction of 4 moles of O2 are required:
4mol O2 * (4mol NH3 / 5mol O2) = 3.2 moles of NH3.
As there are just 1.47 moles, NH3 is limiting reactant
b. Moles NO:
1.47moles NH3 * (4mol NO/4mol NH3) = 1.47mol NO
Mass NO -Molar mass: 30.01g/mol-
1.47mol NO * (30.01g/mol) = 44g of NO
c. Moles H2O:
1.47moles NH3 * (6mol H2O/4mol NH3) = 2.205mol H2O
Mass H2O -Molar mass: 18.01g/mol-
2.205mol H2O * (18.01g/mol) = 40g of H2O
The answer is the 3rd one down I think
