Answer:
144g of H₂O
Explanation:
3NH₄ClO₄(s) + 3Al → Al₂O₃(s) + AlCl₃(s) + 3NO(g) + 6H₂O(g)
From the equation:
3 moles of NH₄ClO₄ produced 6 moles of H₂O
4 moles of NH₄ClO₄ produced ? moles of H₂O
(4 ₓ 6)/3 = 
= 8 moles of H₂O
1 mole of H₂O = (1 × 2) + 16 = 18g (The Relative Molecular mass of H₂O)
8 moles of H₂O = ?
Therefore 8 × 18 = 144g
=144g of H₂O
The correct answer is 1 to the 3rd power
The balanced chemical reaction is:
<span>2C4H10(g)+13O2(g)->10H2O(g)+8CO2(g)
</span>
<span>Calculate the mass of water produced when 1.77 grams of butane reacts with excessive oxygen?
</span>1.77 g C4H10 (1 mol C4H10/58.14 g C4H10) (10 mol H2O / 2 mol C4H10) ( 1.01 g H2O / 1 mol H2O ) = <span>0.15 g H2O
</span><span>Calculate the mass of butane needed to produce 71.6 of carbon dioxide.
</span>71.6 g CO2 (1 mol CO2/ 44.01 g CO2) ( 2 mol C4H10 / 8 mol CO2 ) (58.14 g C4H10 / 1 mol C4H10 ) = 23.65 g C4H10
This is a missing part of your question:
The equilibrium system between sulfur dioxide gas, oxygen gas, and sulfur trioxide gas is given.
So you need the equilibrium balanced equation of SO2, O2, SO3 reaction:
First, we will start with the original equation which is not balanced yet (to understand how we get it):
SO2 + O2 ↔ SO3
Here the number of O atom is not equal at the to sides
So we will start to balance our equation by make the number of O atom equal each other on both sides:
So we will start to put 2SO3 instead of SO3
and put 2SO2 instead of SO2 to balance also the S atom on both sides
So we will get this:
2SO2(g) + O2(g) ↔ 2SO3(g) (This is our equilibrium balanced equation)
know we have a number of O atom equals on each side = 6
and the sulfur equals on each side = 2
