The theoretical yield of acetate is 2607 g. The actual yield of acetate is 1066.8 g. The percentage yield of acetate is 41%.
If 1 mole of vinegar contains 6.02 x 10^23 particles
x moles of vinegar contains 9.02 x 10^24 particles
x = 1 mole x 9.02 x 10^24 /6.02 x 10^23
x = 15 moles of vinegar
The reaction is as follows;
2HC2H3O2 + CaCO3 -----> Ca(C2H3O2)2 + H2O + CO2
Since 2 moles of vinegar reacts with 1 mole of carbonate
x moles of vinegar reacts with 16.5 moles of carbonate
x = 2 moles x 16.5 moles/ 1 mole
x = 33 moles of vinegar
We can see that the vinegar is the reactant in excess hence the carbonate is the limiting reactant.
Theoretical yield = 16.5 moles x 158 g/mol = 2607 g
Actual yield = 6.35 moles x 158 g/mol = 1066.8 g
Percent yield = 1066.8 g/2607 g × 100/1
= 41%
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NF3– 0.94– third
NCl3–0.12– second
NBr3–0.08– first
CF4–1.43– fourth
NBr3—NCl3—NF3—CF4
Lowest. Highest
<span>To find the balanced molecular equation, we need to find the symbols for each element and their charges and then balanced the equation. This comes out to:
Al(OH)3(aq) + 3HBR(aq) ---> AlBr3(aq) + 3H2O(l)
The net ionic equation is adding the charges and separating them into basic components.
Al3+ + 3OH- + 3H+ + 3BR- --> Al3+ + 3Br + H3O+
We can cancel out the aluminum and bromine to get:
3OH-(aq) + 3H+(aq) --> 3H2O(l)</span>
Answer:
corn oil ,2.50 will be receiving
Answer:
0.676 grams of manganese (IV) oxide should be added.
Explanation:
Moles of chlorine gas = n
Volume of the chlorine gas = V = 205 mL = 0.205 L
Pressure of the chlorine gas = 705 Torr = 
1 atm = 760 Torr
Temperature of the chlorine gas = T = 25°C = 25 + 273 K = 298 K
( ideal gs equation)
According to reaction, 1 mole of chlorine gas is obtained from 1 mole of manganese(IV) oxide,then 0.00777 moles of chlorine gas will be obtained from :
of manganese (IV) oxide
Mass of 0.00777 moles of manganese (IV) oxide:
0.00777 mol × 87 g/mol = 0.676 g
0.676 grams of manganese (IV) oxide should be added.
