An example of a monatomic cation would be Cs+. As this is the only cation that consists of a single atom.
Answer: 24.13 g Cu
Explanation:
<u>Given for this question:</u>
M of CuO = 30 g
m of CuO = 79.5 g/mol
Number of moles of CuO = (given mass ÷ molar mass) = (30 ÷ 79.5) mol
= 0.38 mol
The max number of CuO (s) that can be produced by the reaction of excess methane can be solved with this reaction:
CuO(s) + CH4(l) ------> H2O(l) + Cu(s) + CO2(g)
The balanced equation can be obtained by placing coefficients as needed and making sure the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side
4CuO(s) + CH4(l) ----> 2H2O(l) + 4Cu(s) + CO2(g)
From the stoichiometry of the balanced equation:
4 moles of CuO gives 4 moles of Cu
1 mole of CuO gives 1 mol of Cu
0.38 mol of CuO gives 0.38 mol of Cu
Therefore, the grams of Cu that can be produced = 0.38 × molar mass of Cu
= 0.38 × 63.5 g
= 24.13 grams
Therefore, 24.13 grams of copper could be produced by the reaction of 30.0 of copper oxide with excess methane
Answer:
1640 kJ are involved in the reaction
Explanation:
In the reaction:
B₂H₆(g) + 6Cl₂(g) → 2 BCl₃(g) + 6HCl(g)
<em>1 mol of B₂H₆(g) with 6 moles of Cl₂(g) produce 1396 kJ of energy.</em>
<em />
Now if 32.5g of B₂H₆(g) react with excess Cl₂(g), moles involved in reaction are:
If 1 mol produce 1396kJ of energy, 1.175 moles produce:
Thus, <em>1640 kJ are involved in the reaction</em>
First, you have to convert mass to moles by making use of molecular weights [N = 14 g/mol and O = 16 g/mol)
A.
5.6/14 = 0.4 mol N
3.2/16 = 0.2 mol O
mol ratio = 2 mol N/1 mol O => N2O
B.
3.5/14 = 0.25 mol N
8.0/16 = 0.5 mol O
mol ratio = 1 mol N/2 mol O => NO2
C.
1.4/14 = 0.1 mol N
4.0/16 = 0.25 mol O
mol ratio = 2 mol N/5 mol O => N2O5
