Suppose hydrochloric acid reacts with potassium sulfite yielding water, sulfur dioxide, and potassium chloride. Suppose 4 moles
1 answer:
Answer:
128 grams of sulfur dioxide are produced.
Explanation:
Moles of HCl = 4 moles
According to reaction, 2 moles of HCl gives 1 mole of sulfur dioxide gas.
Then 4 moles of HCl will give:
of sulfur dioxide gas.
Mass of sulfur dioxide gas = 2 mol × 64 g/mol = 128 g
128 grams of sulfur dioxide are produced.
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Answer:
a) 2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂
b) Ni(OH)₂
c) KOH
d) 0.927 g
e) K⁺=0.067 M, SO₄²⁻=0.1 M, Ni²⁺=0.067 M
Explanation:
a) The equation is:
2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂ (1)
b) The precipitate formed is Ni(OH)₂
c) The limiting reactant is:
From equation (1) we have that 2 moles of KOH react with 1 mol of NiSO₄, so the number of moles of KOH is:
Hence, the limiting reactant is KOH.
d) The mass of the precipitate formed is:
e) The concentration of the SO₄²⁻, K⁺, and Ni²⁺ ions are:
I hope it helps you!