2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)
65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.
Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)
=0.0282669621 g of O2 left over
Answer:
1.64g
Explanation:
The reaction scheme is given as;
2-bromocyclohexanol --> 1,2-epoxycyclohexane + HBr
From the reaction above,
1 mol of 2-bromocyclohexanol produces 1 mol of 1,2-epoxycyclohexane
3.0 grams of trans-2-bromocyclohexanol.
Molar mass = 179.05 g/mol
Number of moles = mass / molar mass = 3 / 179.05 = 0.016755 mol
This means 0.016755 mol of 1,2-epoxycyclohexane would be produced.
Molar mass = 98.143 g/mol
Theoretical yield = Number of moles * Molar mass
Theoretical yield = 0.016755 * 98.143 ≈ 1.64g
Answer: 0.20 M
Explanation:
According to the dilution law,
where,
= molarity of stock solution = 1.40 M
= volume of stock solution = 72.0 ml
= molarity of diluted solution = m
= volume of diluted solution = 248 ml
Now 124 mL portion of this prepared solution is diluted by adding 133 mL of water.
According to the dilution law,
where,
= molarity of stock solution = 0.41 M
= volume of stock solution = 124 ml
= molarity of diluted solution = m
= volume of diluted solution = (124 +133) ml = 257 ml
Thus the final concentration of the solution is 0.20 M.
Since the oxygen likes to hog all the electrons, it gives the hydrogen molecules a slight positive charge and the itself a slightly negative charge.
Positive attracts negative, so there is some sort of attraction between water molecules, though a weak one.
