Moles of phosphoric acid would be needed : 0.833
<h3>Further explanation</h3>
Given
15 grams of water
Required
moles of phosphoric acid
Solution
Reaction(decomposition) :
H3PO4 -> H2O + HPO3
mol water (H2O :
= mass : MW
= 15 g : 18 g/mol
= 0.833
From the equation, mol ratio H3PO4 = mol H2O = 1 : 1, so mol H3PO4 = 0.833
Answer: So if you had 570 cm of ribbon, then 570%2F8.5=67.05 which means that about 67 students can do the experiment (round down to the nearest whole number).
Explanation: If you had 8.5 cm of ribbon, then only 8.5%2F8.5=1 student can do the experiment. If you had 17 cm of ribbon, then 17%2F8.5=2 students can do the experiment.
The first blank is a compound, second is a mixture
Answer:
There is a production of 11.6 moles of CO₂
Explanation:
The reaction is this:
2C₂H₆(g) + 7O₂(g) ⟶ 4CO₂(g) + 6H₂O(g)
2 moles of ethane reacts with 7 moles of oxygen, to make 4 mol of dioxide and 6 moles of water vapor.
If the oxygen is in excess, we make the calculate with the ethane (limiting reactant)
2 moles of ethane produce 4 moles of dioxide
5.8 moles of ethane produce (5.8 .4)/2 = 11.6 moles
