Answer:
39.3%
Explanation:
CaF2 + H2SO4 --> CaSO4 + 2HF
We must first determine the limiting reactant, the limiting reactant is the reactant that yields the least number of moles of products. The question explicitly says that H2SO4 is in excess so CaF2 is the limiting reactant hence:
For CaF2;
Number of moles reacted= mass/molar mass
Molar mass of CaF2= 78.07 g/mol
Number of moles reacted= 11g/78.07 g/mol = 0.14 moles of Calcium flouride
Since 1 mole of calcium fluoride yields two moles of 2 moles hydrogen fluoride
0.14 moles of calcium fluoride will yield 0.14×2= 0.28 moles of hydrogen fluoride
Mass of hydrogen fluoride formed (theoretical yield) = number of moles× molar mass
Molar mass of hydrogen fluoride= 20.01 g/mol
Mass of HF= 0.28 moles × 20.01 g/mol= 5.6 g ( theoretical yield of HF)
Actual yield of HF was given in the question as 2.2g
% yield of HF= actual yield/ theoretical yield ×100
%yield of HF= 2.2/5.6 ×100
% yield of HF= 39.3%
The pool being cooler at the deep end. Also the air closer to the roof of a house being cooler.
Answer:
Explanation:
We'll assume there is an excess of silver nitrate, so that all 12.0 moles of the magnesium (Mg) will react.
The balanced equation tells us we'll obtain 2 moles of Ag for every 1 mole of magnesium, for a molar ratio of 2/1.
Starting with 12.00 moles Mg, we would therefore hope to find twice that, or 24.00 moles of Ag.
To convert to grams, find the molar mass of Ag from the periodic table.
Ag has a molar mass of 107.9 (to 4 sig figs) grams/mole.
(24.00 moles)*(107.9 grams/mole) = 2590 grams (4 sig figs)
Hands off, it's mine.
Answer:
310.53 g of Cu.
Explanation:
The balanced equation for the reaction is given below:
CuSO₄ + Zn —> ZnSO₄ + Cu
Next, we shall determine the mass of CuSO₄ that reacted and the mass Cu produced from the balanced equation. This can be obtained as follow:
Molar mass of CuSO₄ = 63.5 + 32 + (16×4)
= 63.5 + 32 + 64
= 159.5 g/mol
Mass of CuSO₄ from the balanced equation = 1 × 159.5 = 159.5 g
Molar mass of Cu = 63.5 g/mol
Mass of Cu from the balanced equation = 1 × 63.5 = 63.5 g
Summary:
From the balanced equation above,
159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.
Finally, we shall determine the mass of Cu produced by the reaction of 780 g of CuSO₄. This can be obtained as follow:
From the balanced equation above,
159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.
Therefore, 780 g of CuSO₄ will react to produce = (780 × 63.5)/159.5 = 310.53 g of Cu.
Thus, 310.53 g of Cu were obtained from the reaction.
