Question

section b: word problems(40 points)stoichiometry1. mole-to-mole (synthesis of water)the reaction between hydrogen gas (h₂) and oxygen gas (o₂) produces water (h₂o).- balanced equation:- if you completely react 4.50 moles of oxygen gas (o₂), how many moles of water (h₂o) will be produced? moles of h₂o: ____ mol2. mole-to-mass (combustion of methane)methane gas (ch₄) undergoes combustion, reacting with oxygen (o₂) to produce carbon dioxide (co₂) and water (h₂o).- balanced equation:- if 0.750 moles of methane (ch₄) burns completely, what mass (in grams) of carbon dioxide (co₂) is produced? mass of co₂: ____ gboyles law1. a gas sample contained in a cylinder equipped with a moveable piston occupied 300.0 ml at a pressure of 2.00 atm. what would be the final pressure if the volume were increased to 500.0 ml at constant temperature? (1.20atm)2. a balloon that contains 1.50 l of air at 1.00 atm is taken underwater to a depth at which the pressure is 3.00 atm. calculate the new volume of the balloon. assume that the temperature remains constant. (0.500l)

Explanation:

Stoichiometry

1. Mole-to-Mole (Synthesis of Water)

Step1: Write balanced reaction

$\ce{2H2(g) + O2(g) -> 2H2O(l)}$

Step2: Find mole ratio of $\ce{O2:H2O}$

Mole ratio = $\frac{2\ \text{mol}\ \ce{H2O}}{1\ \text{mol}\ \ce{O2}}$

Step3: Calculate $\ce{H2O}$ moles

$4.50\ \text{mol}\ \ce{O2} \times \frac{2\ \text{mol}\ \ce{H2O}}{1\ \text{mol}\ \ce{O2}}$

Step1: Write balanced reaction

$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$

Step2: Find mole ratio of $\ce{CH4:CO2}$

Mole ratio = $\frac{1\ \text{mol}\ \ce{CO2}}{1\ \text{mol}\ \ce{CH4}}$

Step3: Calculate $\ce{CO2}$ moles

$0.750\ \text{mol}\ \ce{CH4} \times \frac{1\ \text{mol}\ \ce{CO2}}{1\ \text{mol}\ \ce{CH4}} = 0.750\ \text{mol}\ \ce{CO2}$

Step4: Molar mass of $\ce{CO2}$

$M_{\ce{CO2}} = 12.01\ \text{g/mol} + 2\times16.00\ \text{g/mol} = 44.01\ \text{g/mol}$

Step5: Calculate $\ce{CO2}$ mass

$0.750\ \text{mol} \times 44.01\ \text{g/mol}$

Step1: List given values

$P_1=1.20\ \text{atm}, V_1=300.0\ \text{mL}, P_2=2.00\ \text{atm}$

Step2: Apply Boyle's Law ($P_1V_1=P_2V_2$)

Rearrange for $V_2$: $V_2 = \frac{P_1V_1}{P_2}$

Step3: Substitute values

$V_2 = \frac{1.20\ \text{atm} \times 300.0\ \text{mL}}{2.00\ \text{atm}}$

Answer:

9.00 mol

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2. Mole-to-Mass (Combustion of Methane)