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Test: Stoichiometry - JAMB MCQ


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10 Questions MCQ Test Chemistry for JAMB - Test: Stoichiometry

Test: Stoichiometry for JAMB 2024 is part of Chemistry for JAMB preparation. The Test: Stoichiometry questions and answers have been prepared according to the JAMB exam syllabus.The Test: Stoichiometry MCQs are made for JAMB 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Stoichiometry below.
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Test: Stoichiometry - Question 1

Which of the following is not true regarding balanced chemical equations?

Detailed Solution for Test: Stoichiometry - Question 1

A balanced reaction may not have an equal number of molecules on both the sides, because molecules may combine into one or a single molecule may breakdown into two or more. IT can disassociate or undergo double decomposition.

Test: Stoichiometry - Question 2

What is the amount of water produced when 8g of hydrogen is reacted with 32g of oxygen?

Detailed Solution for Test: Stoichiometry - Question 2

The chemical equation of water formation is 2H2 + O2 → 2H2O. Though we have 8g of hydrogen, here oxygen is the limiting reagent. So the only 4g of hydrogen can be used to produce water i.e. 36g of water. That is 2 moles.

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Test: Stoichiometry - Question 3

A and B are two solutions that are mixed. Calculate the resultant solution’s molarity.

Detailed Solution for Test: Stoichiometry - Question 3

When two solutions are combined, the resultant molarity is M1V1 + M2V2/V1 + V2. Consider M1 = 1.3, M2 = 0.8, V1 = 100ml and V2 = 500ml. Now resultant molarity = 1.3(100) + 0.8(500)/100 + 500 = 0.88 mol/L.

Test: Stoichiometry - Question 4

In a container, there are 4 moles of nitrogen, 3 moles of oxygen and 7 moles of hydrogen; find out the mole fraction of oxygen in this reaction.

Detailed Solution for Test: Stoichiometry - Question 4

Mole fraction of a substance is given by the formula: Mole fraction = No. of moles of that substances/No. of total moles of solution. Mole fraction of oxygen here = 3/4 + 3 + 7 = 3/14 = 0.2143.

Test: Stoichiometry - Question 5

What’s the balanced equation of CO2 + H2O → C6H12O6 + O2?

Detailed Solution for Test: Stoichiometry - Question 5

6 CO2 + 6 H2O → C6H12O6 + 6 O2 is the balanced equation because the number of atoms of different elements is the same on both sides. Rest of them do not have an equal number of atoms of different elements on both sides.

Test: Stoichiometry - Question 6

In a particular reaction, one of the reactants limits the number of products formed. That is called as _______

Detailed Solution for Test: Stoichiometry - Question 6

Though the other substances are excess in amount than the required, each and every reactant needs to be in a fixed ratio to attain the desired product. So, thereby, the reactant that limits the quantity of the product formed is called limiting reagent and this reactant gets consumed first completely.

Test: Stoichiometry - Question 7

Which of the given reactions are counted as balanced reactions?

Detailed Solution for Test: Stoichiometry - Question 7

4Al + 3O2 → 2Al2O3 is a balanced reaction because the number of atoms of different elements on both sides is equal. The correctly balanced equations of the rest are 2H2 + O2 → 2H2O; Mg(OH)2 + 2HNO3 → Mg(NO3)2 + 2H2O and N2 + 3H2 → 2NH3.

Test: Stoichiometry - Question 8

Calculate the mass percent of magnesium in the formation of magnesium oxide.

Detailed Solution for Test: Stoichiometry - Question 8

The chemical equation of formation of magnesium oxide is 2 Mg(s) + O2(g) ® 2MgO(s). The formula for calculation of mass percent is (mass of solute/mass of solution) x 100. Therefore 2(24)/2(24+16) = 0.6.

Test: Stoichiometry - Question 9

A solution contains 8 moles of solute and the mass of the solution is 4 kg. What’s the molality of this solution?

Detailed Solution for Test: Stoichiometry - Question 9

Molality of a solution is given y the formula; molality = number of moles of the solute/mass of the solution. So here the number of moles is 8 and the mass of the solution is 4 kg. Molality is 8/4 = 0.5 mol/kg.

Test: Stoichiometry - Question 10

Find the amount of carbon dioxide produced by the combustion of 20g of methane.

Detailed Solution for Test: Stoichiometry - Question 10

The chemical balanced equation for combustion of methane is CH4(g) + 2O2(g) →CO2(g) + 2H2O(g). From the above equation, 1 mole of methane gives 1 mole of carbon dioxide. But 20g of methane = 1.25 moles, therefore it gives 1.25 moles of carbon dioxide = 44(1.5) = 66g.

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