Answer:
11.3 g of H₂O will be produced.
Explanation:
The combustion is:
2C₈H₁₈ + 25O₂→ 16CO₂ + 18H₂O
First of all, we determine the moles of the reactants in order to find out the limiting reactant.
8 g / 114g/mol = 0.0701 moles of octane
37g / 32 g/mol = 1.15 moles of oxygen
The limiting reagent is the octane. Let's see it by this rule of three:
25 moles of oxygen react to 2 moles of octane so
1.15 moles of oxygen will react to ( 1.15 . 2)/ 25 = 0.092 moles of octane.
We do not have enough octane, we need 0.092 moles and we have 0.0701 moles. Now we work with the stoichiometry of the reaction so we make this rule of three:
2 moles of octane produce 18 moles of water
Then 0.0701 moles of octane may produce (0.0701 . 18)/2= 0.631 moles of water.
We convert the moles to mass → 0.631 mol . 18 g/1mol = 11.3 g of H₂O will be produced.
The # 4 is the only significant # in 400.. Trailing 0's r not significant unless there is a decimal.
in 0.7000, there are 4 significant #'s.
this is a dilution equation where 50.0 mL of 1.50 M H₂SO₄ is taken and added to 200 mL of water.
c1v1 = c2v2
where c1 is concentration and v1 is volume of the concentrated solution
and c2 is concentration and v2 is volume of the diluted solution to be prepared
50.0 mL of 1.50 M H₂SO₄ is added to 200 mL of water so the final solution volume is - 200 + 50.0 = 250 mL
substituting these values in the formula
1.50 M x 50.0 mL = C x 250 mL
C = 0.300 M
concentration of the final solution is A) 0.300 M
There is 118 elements<span> in the periodic table</span>
Answer:
7.25 x 10^47
Explanation:
1.204 x 10^24 moles*6.022 x 10^23 avogadro's number= 7.25 x 10^
