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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Answer:
The concentration of NaCl = 0.3374 M
Explanation:
Given :
Molarity of AgNO₃ = 0.2503 M
Volume of AgNO₃ = 20.22 mL
The conversion of mL into L is shown below:
Thus, volume of the solution = 20.22×10⁻³ L
Molarity of a solution is the number of moles of solute present in 1 L of the solution.
The formula can be written for the calculation of moles as:
Thus,
The chemical reaction taking place:
According to reaction stoichiometry:
<u>1 mole</u> of AgNO₃ reacts with <u>1 mole</u> of NaCl
Thus,
5.0611×10⁻³ moles of AgNO₃ reacts with 5.0611×10⁻³ moles of NaCl
Thus, moles of NaCl required = 5.0611×10⁻³ moles
Volume of NaCl required = 15.00 mL
The conversion of mL into L is shown below:
Thus, volume of the solution = 15.00×10⁻³ L
Applying in the formula of molarity as:
<u>Thus, the concentration of NaCl = 0.3374 M</u>