5.00 mL of a 2.00 M Ca(NO3)2 are added to 25.0 mL of 2.00 M sodium sulfate. 0.975 g of a solid product forms. What is the percent yield of the solid? The answer is a percent. Do not include units in your answer. Scientific notation is optional. To write scientific notation in Canvas, type "e" to replace "x 10". For example, "1 x 1014" is written "1e+14" in Canvas.

To calculate the percent yield of the solid, we need to determine the theoretical yield and the actual yield. Let's start by balancing the chemical equation for the reaction:

Ca(NO3)2 + Na2SO4 → CaSO4 + 2NaNO3

The balanced equation shows that 1 mole of Ca(NO3)2 reacts with 1 mole of Na2SO4 to produce 1 mole of CaSO4. From the given information, we know that the initial volume of Ca(NO3)2 is 5.00 mL and the initial volume of Na2SO4 is 25.0 mL. Both solutions have a concentration of 2.00 M.

Using the formula:

Moles = Volume (L) x Concentration (M)

We can calculate the moles of Ca(NO3)2 and Na2SO4:

Moles of Ca(NO3)2 = 5.00 mL x 0.00200 M = 0.0100 moles

Moles of Na2SO4 = 25.0 mL x 0.00200 M = 0.0500 moles

Since the stoichiometry of the reaction is 1:1, the moles of CaSO4 formed will be equal to the moles of Ca(NO3)2 used.

Therefore, the theoretical yield of CaSO4 is 0.0100 moles.

Given that the actual yield of the solid is 0.975 g, we can convert this to moles using the molar mass of CaSO4:

Molar mass of CaSO4 = 40.08 g/mol + 32.06 g/mol + (4 x 16.00 g/mol) = 136.14 g/mol

Moles of CaSO4 = 0.975 g / 136.14 g/mol = 0.00716 moles

Now we can calculate the percent yield:

Percent Yield = (Actual Yield / Theoretical Yield) x 100%

Percent Yield = (0.00716 moles / 0.0100 moles) x 100% = 71.6%

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