How many milliliters of 3.60 {\rm M} {\rm HCl} are required to react with 140 {\rm mL} of 1.60 {\rm M}{\rm Al}({\rm OH})_{3}?
I don't understand your hieroglyphics but here is the answer.
3HCl + Al(OH)3 ==> 3H2O + AlCl3
moles Al(OH)3 = M x L = ??
Using the coefficients in the balanced equation, convert moles Al(OH)3 to moles HCl. That will be 3*moles Al(OH)3 = moles HCl.
Then M HCl = moles HCl/L HCl and convert to mL.
What is the mass, in kilograms, of the air in a room that measures 14.0 \times\; 14.0 \times\; 8.0 {\rm{ft}}?
To find the number of milliliters of 3.60 M HCl needed to react with 140 mL of 1.60 M Al(OH)3, we can use the concept of stoichiometry and the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between HCl and Al(OH)3 is:
3 HCl + Al(OH)3 -> AlCl3 + 3 H2O
From the balanced equation, we can see that the stoichiometric ratio between HCl and Al(OH)3 is 3:1. This means that for every 3 moles of HCl, 1 mole of Al(OH)3 is required.
Step 1:
First, we need to calculate the number of moles of Al(OH)3 present in 140 mL of 1.60 M Al(OH)3. We can use the formula:
moles = concentration (M) × volume (L)
Given:
Concentration of Al(OH)3 = 1.60 M
Volume of Al(OH)3 = 140 mL
Converting mL to L:
Volume of Al(OH)3 = 140 mL ÷ 1000 = 0.14 L
Using the formula:
moles of Al(OH)3 = 1.60 M × 0.14 L = 0.224 moles
Step 2:
Using the stoichiometric ratio from the balanced equation, we can determine the number of moles of HCl required to react with the calculated moles of Al(OH)3.
Since the stoichiometric ratio is 3:1, the number of moles of HCl required would be 3 times the number of moles of Al(OH)3.
moles of HCl required = 3 × 0.224 moles = 0.672 moles
Step 3:
Now, we need to find the volume of 3.60 M HCl needed to contain 0.672 moles of HCl. We can use the formula:
volume (L) = moles ÷ concentration (M)
Given:
Concentration of HCl = 3.60 M
Moles of HCl = 0.672 moles
Using the formula:
volume of HCl = 0.672 moles ÷ 3.60 M = 0.1867 L
Converting L to mL:
volume of HCl = 0.1867 L × 1000 = 186.7 mL
Therefore, approximately 186.7 mL of 3.60 M HCl is required to react with 140 mL of 1.60 M Al(OH)3.