Azane
And it's Molar Mass : 17.031 g/mol
Answer:
0.478 M
Explanation:
Let's consider the neutralization reaction between KOH and H₂SO₄.
2 KOH + H₂SO₄ → K₂SO₄ + 2 H₂O
12.7 mL of 1.50 M H₂SO₄ react. The reacting moles of H₂SO₄ are:
0.0127 L × 1.50 mol/L = 0.0191 mol
The molar ratio of KOH to H₂SO₄ is 2:1. The reacting moles of KOH are 2 × 0.0191 mol = 0.0382 mol
0.0382 moles of KOH are in 80.0 mL. The molarity of KOH is:
M = 0.0382 mol/0.0800 L = 0.478 M
The solution to your problem is as follows:
<span>2 KClO3 → 2 KCl + 3 O2
</span><span>MW of KCL = 39.1 + 35.35 = 74.6 g/mole
62.6/74.6 = 0.839 moles KCl produced
0.839 moles KCl x 3O2/2KCl x 32g/mole O2 = 40.27g of O2 was produced
</span>
Therefore, there are <span>40.27g of O2 produced during the reaction.
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I hope my answer has come to your help. Thank you for posting your question here in Brainly.
Density will be mass/volume= 0.2/500 = 0.0004
