Mass percentage of a solution is the amount of solute present in 100 g of the solution.
Given data:
Mass of solute H2SO4 = 571.3 g
Volume of the solution = 1 lit = 1000 ml
Density of solution = 1.329 g/cm3 = 1.329 g/ml
Calculations:
Mass of the given volume of solution = 1.329 g * 1000 ml/1 ml = 1329 g
Therefore we have:
571.3 g of H2SO4 in 1329 g of the solution
Hence, the amount of H2SO4 in 100 g of solution= 571.3 *100/1329 = 42.987
Mass percentage of H2SO4 (%w/w) is 42.99 %
The balanced equation that illustrates the reaction is:
2C4H6 + 11O2 ......> 8CO2 + 6H2O
number of moles = mass / molar mass
number of moles of oxygen = 2.1 / 32 = 0.065625 moles
Now, from the balanced equation, we can note that:
11 moles of oxygen are required to produce 6 moles of water.
Therefore:
0.065625 moles of oxygen will produce:
(0.065625*6) / 11 = 0.03579 moles of water
number of moles = mass / molar mass
mass = number of moles * molar mass
mass of water = 0.03579 * 18 = 0.644 grams
Answer:
The mass of water is 36 g.
Explanation:
Mass of hydrogen = 4 g
Mass of water = ?
Solution:
First of all we will write the balance chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of hydrogen = mass / molar mass
Number of moles of hydrogen = 4 g/ 2 g/mol
Number of moles of hydrogen = 2 mol
Now we compare the moles of water with hydrogen from balance chemical equation.
H₂ : H₂O
2 : 2
Mass of water = moles × molar mass
Mass of water = 2 mol × 18 g/mol
Mass of water = 36 g
If the water oxygen is in excess than mass of water would be 36 g.
Answer:
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I think it's D, because theoretical yield is like, the yield you'd get if 100% of the reactants formed to make product. Well that's how I think of it, but it has something to do with limiting reagents and stuff. Sorry this isn't a really detailed explanation.
