Explanation:
Let the mass of isoamyl acetate be 100g.
Moles of Carbon = 60.58/12 = 5.048mol
Moles of Hydrogen = 7.07/1 = 7.07mol
Moles of Oxygen = 32.28/16 = 2.018mol
Mole Ratio of C : H : O
= 5.048 : 7.07 : 2.018
= 5 : 7 : 2.
Hence the empirical formula of isoamyl acetate is C5H7O2.
<span>In a sample of solid Ba(NO3)2 the ratio of barium ions to nitrate ions is would be one is to 2 or 1:2. Barium ion has a formal charge of positive two which means that it needs two ions which has a formal charge of negative one or 1 ion with the formal charge of negative two. However, for this case, it is bonded to a nitrate ion which has a formal charge of negative one. Therefore, it needs two nitrate ions so that for every 1 atom of barium ion, we need two ions of nitrate ions.</span>
Answer:
2.067 L ≅ 2.07 L.
Explanation:
- The balanced equation for the mentioned reaction is:
<em>CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g),</em>
It is clear that 1.0 mole of CS₂ react with 3.0 mole of O₂ to produce 1.0 mole of CO₂ and 2.0 moles of SO₂.
- At STP, 3.6 L of H₂ reacts with (?? L) of oxygen gas:
It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.
<u><em>using cross multiplication:</em></u>
1.0 mol of O₂ represents → 22.4 L.
??? mol of O₂ represents → 3.1 L.
∴ 3.1 L of O₂ represents = (1.0 mol)(3.1 L)/(22.4 L) = 0.1384 mol.
- To find the no. of moles of SO₂ produced from 3.1 liters (0.1384 mol) of hydrogen:
<u><em>Using cross multiplication:</em></u>
3.0 mol of O₂ produce → 2.0 mol of SO₂, from stichiometry.
0.1384 mol of O₂ produce → ??? mol of SO₂.
∴ The no. of moles of SO₂ = (2.0 mol)(0.1384 mol)/(3.0 mol) = 0.09227 mol.
- Again, using cross multiplication:
1.0 mol of SO₂ represents → 22.4 L, at STP.
0.09227 mol of SO₂ represents → ??? L.
∴ The no. of liters of SO₂ will be produced = (0.09227 mol)(22.4 L)/(1.0 mol) = 2.067 L ≅ 2.07 L.
I think you divide something
OK in the case of hydrazine 14 grams of nitrogen combine with 2 gram of hydrogen and with ammonia 14 grams combine with 3 grams of hydrogen.
Ratio 2:3
