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:
Option b. 0.048 M
Explanation:
We have the molecular weight and the mass, from sulcralfate.
Let's convert the mass in g, to moles
1 g . 1 mol / 2087 g = 4.79×10⁻⁴ moles.
Molarity is mol /L
Let's convert the volume of solution in L
10 mL . 1L/1000 mL = 0.01 L
4.79×10⁻⁴ mol / 0.01 L = 0.048 mol/L
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