how many grams of barium sulfate are produced if 10.0 grams of barium chloride are reacted is excess sulfuric acid
1 answer:
Answer: 11.2 grams
Explanation:
The balanced chemical equation for reaction is:
To calculate the moles, we use the equation:
a) moles of
According to stoichiometry :
As sulfuric acid is in excess , the limiting reagent is barium chloride as it limits the formation of product.
1 mole of produces= 1 mole of
Thus 0.0480 moles of require=
of
Mass of
Thus 11.2 g of barium sulfate are produced if 10.0 grams of barium chloride are reacted is excess sulfuric acid
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27.7g
Explanation:
Given parameters:
Number of moles of oxygen = 3.15moles
Unknown:
Mass of reacting propane = ?
Solution:
Equation of the reaction:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
We work from the known to the unknown parameters.
The known in this expression is the number of moles of oxygen.
In the reaction:
5 moles of oxygen gas reacts with 1 mole of propane
3.15 mole of oxgen gas will react with = 0.63moles
Number of moles of reacting propane gas is 0.63moles
To find the mass;
mass = number of moles x molar mass = 0.63 x molar mass
molar mass of propane = (3 x 12) + (8x1) = 44g/mol
mass of propane = 0.63 x 44 = 27.7g
Learn more:
Combustion reaction brainly.com/question/6126420
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