The answer would be D 0.30m
Reasoning Hydroxide,NaOH is 15 ml base 35ml of .017 which is HCI
Answer:
d. Copper (II) sulfate
Explanation:
Given data:
Mass of Al = 1.25 g
Mass of CuSO₄ = 3.28 g
What is limiting reactant = ?
Solution:
Chemical equation:
2Al + 3CuSO₄ → Al₂ (SO₄)₃ + 3Cu
Number of moles of Al:
Number of moles = mass/molar mass
Number of moles = 1.25 g/ 27 g/mol
Number of moles = 0.05 mol
Number of moles of CuSO₄:
Number of moles = mass/molar mass
Number of moles = 3.28 g/ 159.6 g/mol
Number of moles = 0.02 mol
now we will compare the moles of reactant with product.
Al : Al₂ (SO₄)₃
2 : 1
0.05 : 1/2×0.05=0.025 mol
Al : Cu
2 : 3
0.05 : 3/2×0.05 = 0.075 mol
CuSO₄ : Al₂ (SO₄)₃
3 : 1
0.02 : 1/3×0.02=0.007 mol
CuSO₄ : Cu
3 : 3
0.02 : 0.02
Less number of moles of reactants are produced by CuSO₄ thus it will act as limiting reactant.
Answer:
B
Explanation:
The last 2 answers can immediately be cancelled out as they're not combustion of ethanol, combustion in this context is not a reversable reaction so cannot be referenced in this way.
The answer is determined to be B just by simply counting the numbers of each of the elements on the reactants (left) side and the products (right) side and ensuring they're the same
In A on the reactants side there is 2C 6H 3O and on the products side there is 1C 3O and 2H its not equal so it is not correct
In B on the reactants side there is 2C 6H 7O and on the products side there is 2C 6H 7O its equal so this answer is correct
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