Answer:
D no. is the answer of your question
<u>Answer:</u> The mass of iron in the ore is 10.9 g
<u>Explanation:</u>
We are given:
Mass of iron (III) oxide = 15.6 g
We know that:
Molar mass of Iron (III) oxide = 159.69 g/mol
Molar mass of iron atom = 55.85 g/mol
As, all the iron in the ore is converted to iron (III) oxide. So, the mass of iron in iron (III) oxide will be equal to the mass of iron present in the ore.
To calculate the mass of iron in given mass of iron (III) oxide, we apply unitary method:
In 159.69 g of iron (III) oxide, mass of iron present is 
So, in 15.6 g of iron (III) oxide, mass of iron present will be = 
Hence, the mass of iron in the ore is 10.9 g
Answer: 27 is A and 28 is C.
Explanation: I’ll explain 28 but not 27 because that’s just definitions.
In CuSO4 there is a Cu, an S, and 4 O molecules. Add them up you get 6.
Answer:
2.61 g of NO will be formed
The limiting reagent is the O₂
Explanation:
The reaction is:
4NH₃ + 5O₂ → 4NO + 6H₂O
We convert the mass of the reactants to moles:
3.25g / 17 g/mol = 0.191 moles of NH₃
3.50g / 32 g/mol =0.109 moles of O₂
Let's determine the limiting reactant by stoichiometry:
4 moles of ammonia react with 5 moles of oxygen
Then, 0.191 moles of ammonia will react with (0.191 . 5) / 4 = 0.238 moles of oxygen. We only have 0.109 moles of O₂ and we need 0.238, so as the oxygen is not enough, this is the limiting reagent
Ratio with NO is 5:4
5 moles of oxygen produce 4 moles of NO
0.109 moles will produce (0.109 . 4)/ 5 = 0.0872 moles of NO
We convert the moles to mass, to get the answer
0.0872 mol . 30g / 1 mol = 2.61 g
