I have 3 questions I need help with
1. mix 120ml of 0.320M silver nitrate with 40ml of 0.320M potasium chromate. What mass of silver chromate is produced?
2. Calculate the mass of white solid CaCO3 that forms when 98ml of a 0.1M Ca(NO3)2 solution is mixed with 54ml of a 2.5M Na2CO3 solution.
3. Aluminum metal react with an excess HCl solution to produce aluminum cloride and hydrogen gas. If 46g of aluminum is reacted, what mass of hydrogen is produced?
Any help is much appreciated, Thank you in advance :)
Of course! I'll walk you through each question step by step.
1. To find the mass of silver chromate produced, we need to determine the limiting reagent between silver nitrate (AgNO3) and potassium chromate (K2CrO4).
First, let's calculate the number of moles of AgNO3 and K2CrO4 in each solution:
- Moles of AgNO3 = Volume (in L) x Molarity = 0.120 L x 0.320 mol/L = 0.0384 mol
- Moles of K2CrO4 = Volume (in L) x Molarity = 0.040 L x 0.320 mol/L = 0.0128 mol
Now, we need to determine the limiting reagent. Since the stoichiometric ratio between AgNO3 and K2CrO4 is 2:1 (from the balanced equation AgNO3 + K2CrO4 → Ag2CrO4 + 2KNO3), we can see that the limiting reagent is K2CrO4 because it has a smaller number of moles.
Next, we calculate the moles of Ag2CrO4 formed based on the limiting reagent:
- Moles of Ag2CrO4 = 0.0128 mol K2CrO4 x (1 mol Ag2CrO4 / 1 mol K2CrO4) = 0.0128 mol
To find the mass of Ag2CrO4 produced, we need to multiply the moles by its molar mass (331.7 g/mol, from the periodic table):
- Mass of Ag2CrO4 = 0.0128 mol x 331.7 g/mol ≈ 4.24 g
Therefore, approximately 4.24 grams of silver chromate are produced in this reaction.
2. Similarly, in this question, we need to determine the limiting reagent. Let's calculate the number of moles for both reactants:
- Moles of Ca(NO3)2 = Volume (in L) x Molarity = 0.098 L x 0.1 mol/L = 0.0098 mol
- Moles of Na2CO3 = Volume (in L) x Molarity = 0.054 L x 2.5 mol/L = 0.135 mol
Again, the stoichiometric ratio between Ca(NO3)2 and Na2CO3 is 1:1 (from the balanced equation Ca(NO3)2 + Na2CO3 → CaCO3 + 2NaNO3). Since both have an equal number of moles, neither is in excess or limiting. This means that both reactants will completely react with each other.
To find the mass of CaCO3 formed, we multiply the moles by its molar mass (100.1 g/mol, from the periodic table):
- Mass of CaCO3 = 0.0098 mol x 100.1 g/mol = 0.98 g
Therefore, 0.98 grams of white solid CaCO3 is produced.
3. In this question, the balanced chemical equation is: 2Al + 6HCl → 2AlCl3 + 3H2. From this equation, we can see that the stoichiometric ratio between aluminum (Al) and hydrogen gas (H2) is 2:3.
First, calculate the number of moles of aluminum:
- Moles of Al = Mass / Molar mass = 46 g / 26.98 g/mol = 1.704 mol
Using the stoichiometric ratio, we can determine the number of moles of hydrogen gas produced:
- Moles of H2 = 1.704 mol Al x (3 mol H2 / 2 mol Al) = 2.556 mol
Finally, to find the mass of hydrogen gas produced, we multiply the moles by its molar mass (2 g/mol, from the periodic table):
- Mass of H2 = 2.556 mol x 2 g/mol = 5.112 g
Therefore, approximately 5.112 grams of hydrogen gas are produced.