From the mole ratio of the reaction as given in the equation of the reaction, the mass of calcium chloride that can be produced from 5.59 mol of hydrochloric acid is 310.245 g.
<h3>What mass of calcium chloride can be produced from 5.59 mol of hydrochloric acid?</h3>
The mass of calcium chloride that can be produced from 5.59 mol of hydrochloric acid is determined from the equation of the reaction.
The equation of the reaction is given below:
Ca²⁺ (aq) + 2 HCl (aq) ---> CaCl₂ (s) + H₂ (g)
From the equation of the reaction, the mole ratio of HCL and calcium chloride is 2 : 1
Therefore, moles of calcium chloride that can be produced will be:
The moles of calcium chloride = 5.59 moles * 1/2
The moles of calcium chloride = 2.795 moles
The mass of calcium chloride produced = moles * molar mass
Molar mass of calcium chloride = 111 g/mol
Mass of calcium chloride produced = 2.795 * 111
Mass of calcium chloride produced = 310.245 g
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Answer:
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- <u>b. See the description below</u>
Explanation:
<u><em>a. Volume of 0.400 M CuSO₄(aq) required for the preparation</em></u>
In dissolutions, since the number of moles of solute is constant, the equation is:
Substitute and solve for V₁
<u><em>b. Briefly describe the essential steps to most accurately prepare the 0.150 M CuSO₄(aq) from the 0.400 M CuSO₄(aq)</em></u>
You will use the stock solution, the funnel, the buret, and distilled water.
i) Using the funnel, fill in the buret with with 50 ml of the stock solution, i.e. the 100. ml of 0.400 M CuSO₄(aq) solution.
ii) Pour 37.5 ml of the stock solution from the burete into the volumetric flask.
iii) Carefully add disitlled water to the 37.5ml of the stock solution in the volumetric flask until the mark (50 ml) on the volumetric flask.
iv) Put the stopper and rotate the volumetric flask to homegenize the solution.
2)
formula equation: Pb(NO₃)₂(aq) + 2KI(aq) ----> PbI₂(s) + 2KNO₃(aq)
total ionic equation: Pb⁺² + 2NO₃⁻¹ + 2K⁺¹ + 2I⁻¹----> PbI₂(s) + 2K⁺¹ + 2NO₃⁻¹
net ionic equation: Pb⁺² + 2 I⁻¹----> PbI₂(s)
3)
formula equation: Zn(NO₃)₂(aq) + K₂CO₃(aq) -----> ZnCO₃(s) + 2KNO₃(aq)
total ionic equation: Zn⁺² + 2NO₃⁻¹ + 2K⁺¹ + CO₃⁻² ---> ZnCO₃ (s) + 2K⁺¹ + 2NO₃⁻¹
net ionic equation: Zn⁺² + CO₃⁻² ----> ZnCO₃ (s)
note: if I did not specify the state of the molecule in the reaction, you can assume they are aqueous unless state otherwise.
