The formula for mole is
n= Mass/Mol mass
Mol Mass: S=32
O2= 16(2)
—————
64 g/mol
N= 17.50 g
————— (cancel both g)
64 g/mol
= 0.27 mol is the answer
Oxygen gas produced : 0.7 g
<h3>Further explanation</h3>
Given
10.0 grams HgO
9.3 grams Hg
Required
Oxygen gas produced
Solution
Reaction⇒Decomposition
2HgO(s)⇒2Hg(l)+O₂(g)
Conservation of mass applies to a closed system, where the masses before and after the reaction are the same
mass of reactants = mass of products
mass HgO = mass Hg + mass O₂
10 g = 9.3 g + mass O₂
mass O₂ = 0.7 g
B) 40%
The balanced equation indicates that for every 3 moles of H2 used, 2 moles of NH3 will be produced. So the reaction if it had 100% yield would produce (2.00 / 3) * 2 = 1.333333333 moles of NH3. But only 0.54 moles were produced. So the percent yield is 0.54 / 1.3333 = 0.405 = 40.5%. This is a close enough match to option "b" to be considered correct.
It’s A.it’s properties
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