One way of expressing concentration is by percent. It may be on the basis of mass, mole or volume. Percent is expressed as the amount of solute per amount of the solution. For this case, we are given the percent by mass. In order to solve the amount of solute, we multiply the percent with the amount of the solution.
Mass of solute = percent by mass x mass solution
Mass of solute = 0.0350 x 2.50 x10^2 = 8.75 grams of solute
The balanced equation for the above reaction is as follows;
2S + 3O₂ --> 2SO₃
Stoichiometry of O₂ to SO₃ is 3:2
O₂ is the limiting reactant and S is provided in excess. since O₂ is the limiting reactant, the whole amount is consumed in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of O₂ moles reacted- 4 g / 32 g/mol = 0.125 mol
3 mol of O₂ forms 2 mol of SO₃
therefore when 0.125 mol of O₂ reacts number of SO₃ moles - 2/3 x 0.125 mol
Number of SO₃ moles formed - 0.0833 mol
Answer is 4) 0.08 mol
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