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NECO SSCE Stoichiometry of Chemical Reactions — practice questions — page 3

Explore 79 free Stoichiometry of Chemical Reactions questions in NECO SSCE Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Stoichiometry of Chemical Reactions questions and exercises with answers

Showing questions 61–79 of 79.

Reacting 1.274 g of CuSO4 with excess zinc gives 0.392 g of copper, against a theoretical 0.5072 g. What is the percent yield?Read the answer and worked solution →A student isolated 25 g of a compound from a procedure that should theoretically yield 81 g. What is the percent yield?Read the answer and worked solution →Reacting 3.0 mol of H2 with 2.0 mol of I2 gave 1.0 mol of HI. What is the percent yield for H2 + I2 giving 2HI?Read the answer and worked solution →For 2Cr + 2H3PO4 giving 2CrPO4 + 3H2, which is limiting when 0.50 mol Cr meets 0.75 mol H3PO4?Read the answer and worked solution →How is the atom economy of a synthesis defined?Read the answer and worked solution →What is quantitative analysis?Read the answer and worked solution →What is a buret used for in titration analysis?Read the answer and worked solution →A 50.00 mL sample of HCl needs 35.23 mL of 0.250 M NaOH to reach the end point. What is the HCl molarity?Read the answer and worked solution →A 20.00 mL oxalic acid sample needs 23.24 mL of 0.09113 M KMnO4, reacting as 2MnO4- with 5H2C2O4. What is the acid molarity?Read the answer and worked solution →What defines a gravimetric analysis?Read the answer and worked solution →Drying a sample by controlled heating to find its water content is an example of which method?Read the answer and worked solution →A 0.4550 g mixture treated with excess Ba(NO3)2 gives 0.6168 g of BaSO4 (233.43 g/mol). What is the mass percent of MgSO4 (120.37 g/mol)?Read the answer and worked solution →Burning 0.00126 g of polyethylene gives 0.00394 g CO2 (44.01 g/mol) and 0.00161 g H2O (18.02 g/mol). What is its empirical formula?Read the answer and worked solution →In the old vinegar test, potassium carbonate was added until bubbling stopped. What gas caused the bubbling?Read the answer and worked solution →Balance the splitting of waterRead the answer and worked solution →Order the mass to mass routeRead the answer and worked solution →Fill in the percent yield formulaRead the answer and worked solution →Knock out the false limiting claimsRead the answer and worked solution →Pop only the soluble compoundsRead the answer and worked solution →

Worked practice questions

Reacting 1.274 g of CuSO4 with excess zinc gives 0.392 g of copper, against a theoretical 0.5072 g. What is the percent yield?

  1. 48.3%
  2. 30.8%
  3. 129%
  4. 77.3%
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Correct answer: D. 77.3%

1. Zinc is in excess, so copper sulfate is the limiting reactant and fixes the theoretical yield. 2. Converting 1.274 g of CuSO4 through 159.62 g/mol and then 63.55 g/mol gives 0.5072 g of copper. 3. The actual yield collected was 0.392 g of copper. 4. Percent yield is 0.392 divided by 0.5072, then multiplied by 100. 5. That division gives 0.7729, so the percent yield is 77.3%. 6. Dividing the other way round gives 129%, which is impossible and flags the reversed formula. 7. Rounding only at the end keeps the third significant figure trustworthy. _Source: OpenStax Chemistry (CC BY 4.0), Ch 4 "Stoichiometry of Chemical Reactions", section 4.4 Reaction Yields_

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A student isolated 25 g of a compound from a procedure that should theoretically yield 81 g. What is the percent yield?

  1. 25%
  2. 31%
  3. 56%
  4. 324%
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Correct answer: B. 31%

1. Percent yield is the actual yield divided by the theoretical yield, times 100. 2. Here the actual yield is 25 g and the theoretical yield is 81 g. 3. Dividing gives 25 divided by 81, which is 0.3086. 4. Multiplying by 100 gives 30.86%, that is about 31%. 5. Subtracting the masses to get 56 g is a different quantity and not a percent yield. _Source: OpenStax Chemistry (CC BY 4.0), Ch 4 "Stoichiometry of Chemical Reactions", section 4.4 Reaction Yields_

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Reacting 3.0 mol of H2 with 2.0 mol of I2 gave 1.0 mol of HI. What is the percent yield for H2 + I2 giving 2HI?

  1. 17%
  2. 33%
  3. 50%
  4. 25%
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Correct answer: D. 25%

1. The equation needs hydrogen and iodine in a 1 to 1 mole ratio. 2. With 3.0 mol H2 and 2.0 mol I2, iodine is the limiting reactant. 3. Complete reaction of 2.0 mol I2 would give 2 times 2.0, that is 4.0 mol HI. 4. Hydrogen alone could have given 6.0 mol HI, confirming that iodine caps the yield. 5. So the theoretical yield is 4.0 mol HI and the actual yield is 1.0 mol HI. 6. Percent yield is 1.0 divided by 4.0, times 100, which is 25%. 7. Using 3.0 mol H2 as the basis would wrongly give about 17%, which is the limiting-reactant trap. _Source: OpenStax Chemistry (CC BY 4.0), Ch 4 "Stoichiometry of Chemical Reactions", section 4.4 Reaction Yields_

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