Find the number of moles of sodium you have:
<span>n = m/M where m is your 20g of sodium and M is 22.99 g/mol. </span>
<span>Look at the stoichiometry of the equation - it's 2:2 when you are producing NaOH. So if you took 1 mole of Na, it'd produce 1 mole of NaOH (as the ratio is equal). </span>
<span>That means that your moles of sodium is equal to the moles of NaOH produced. Use the molar mass of NaOH - which is 39.998 g/mol along with your calculated number of moles to get the mass (the formula rearranges to m = nM). </span>
<span>This figure is the theoretical yield - what you would get if every last mole of sodium was converted into NaOH. </span>
<span>What you get in practice is the experminetal yield, and the percentage yield is the experimental yield divided by the theoretical yield - and then multiplied by 100%.</span>
Answer:
93%
Explanation:
First write the balanced reaction
C3H8 + 5O2 --> 3CO2 +4H2O
2nd, find the theoretical yield
2L C3H8 (1mol/22.4L) (3 mol CO2/1 molC3H8) (22.4 L/ 1mol) = 6L
3rd, Find percent yield
%yield = (actual / theoretical) x100 = (5.7/6) x 100 = 93%
Answer:
This is an acid-base reaction (neutralization): H 2SO 4 is an acid, Ca(OH) 2 is a base.
...
1 H2SO4 + Ca(OH)2 → H2O + CaSO4
2 H2SO4 + Ca(OH)2 → CaSO4 + HOH
3 H2SO4 + Ca(OH)2 → H2O + CaOH2SO4
4 H2SO4 + Ca(OH)2 → H2O + Ca(HSO4)2
5 H2SO4 + Ca(OH)2 → H2O + (CaOH)2SO4
Explanation:
Answer:
The correct answer is 4 substracting hydride ions from olefins.
Explanation:
The olefins are the unsaturated hydrocarbon also called alkenes that contain one double bond present between 2 carbon atoms.
On the other hand a carbenium ion is formed when a hydrocarbon loses a hydride ion to form a positively charged product.
For example allyl cation
CH2=CH-CH2+
The allyl cation is a carbenium ion formed by the removal of H- ion or hydride ion from the alkene propene.
