Hello! I am having a little difficulty answering these two questions on my homework, I was wondering if anyone could please help me out! Thank you!
1) Consider the following balanced chemical equation for the neutralization reaction of calcium hydroxide with hydrochloric acid to form calcium chloride and water.
Ca(OH)2(s) + 2HCl(aq) --> CaCl2(aq) + 2H2O(l)
How many mL of 0.175 M HCl are required to react with 46.5 mL of 0.125 M calcium hydroxide?
*Use a mole/mole ratio from the equation, and molarities.
2)Consider the following balanced chemical equation for the single replacement reaction of calcium metal with aqueous HCl to form calcium chloride and hydrogen gas.
Ca(s) + 2HCl(aq) --> CaCl2(aq) + H2(g)
How many mL of 3.31 M HCl are required to react with 2.94 g of calcium?
*Use the chemical equation to write a mole/mole ratio
We need 2 moles of HCl for every mole of calcium hydroxide
we have 46.5 *10^-3 * .125 = 5.81 *10^-3 moles of calcium hydroxide
so we need
11.6 *10^-3 moles of HCl
.175 moles/liter * x liters = 11.6*10^-3 moles
x = 66.4 * 10^-3 liters = 66.4 mL
the second one is very similar except you need to know how many moles of Calcium are 2.94 grams
atomic mass of calcium = 40 grams/mole approx
so
2.94 g /40 g/mole = .0735 moles of Ca
so you will need twice that many moles of HCl
Hello! I'd be happy to help you with these questions. Let's tackle them one at a time.
1) To solve this problem, we can use the concept of mole-mole ratios. We're given the molarities of the solutions and the volumes in milliliters. Here are the steps to find the answer:
a) Convert the given volumes into liters:
- 46.5 mL of 0.125 M calcium hydroxide = 46.5 mL * (1 L / 1000 mL) = 0.0465 L
b) Use the mole-mole ratios from the balanced chemical equation to calculate the moles of HCl and Ca(OH)2:
- From the equation: 1 mole of Ca(OH)2 reacts with 2 moles of HCl
- Moles of Ca(OH)2 = (0.0465 L) * (0.125 mol/L) = 0.00581 mol
c) Use the mole-mole ratio to calculate the moles of HCl needed:
- Moles of HCl = 0.00581 mol * (2 mol HCl / 1 mol Ca(OH)2) = 0.0116 mol
d) Convert the moles of HCl to volume using its molarity:
- Volume of HCl = (0.0116 mol) / (0.175 mol/L) = 0.0663 L = 66.3 mL
Therefore, you would need 66.3 mL of 0.175 M HCl to react with 46.5 mL of 0.125 M calcium hydroxide.
2) Similar to the first question, we can use the mole-mole ratios to solve this problem. Here are the steps:
a) Convert the given mass of calcium into moles:
- Moles of calcium = (2.94 g) * (1 mol Ca / 40.08 g) = 0.0731 mol
b) Use the mole-mole ratio from the balanced chemical equation to calculate the moles of HCl needed:
- From the equation: 1 mole of Ca reacts with 2 moles of HCl
- Moles of HCl = 0.0731 mol * (2 mol HCl / 1 mol Ca) = 0.146 mol
c) Convert the moles of HCl to volume using its molarity:
- Volume of HCl = (0.146 mol) / (3.31 mol/L) = 0.044 L = 44 mL
Therefore, you would need 44 mL of 3.31 M HCl to react with 2.94 g of calcium.
Remember, the key to solving stoichiometry problems like these is using the mole-mole ratios from the balanced chemical equation and making sure to convert between moles and desired units (such as liters or milliliters) using the given molarities. I hope this explanation helps you understand how to solve these types of problems! Let me know if you have any further questions.