Zinc reacts with hydrochloric acid according to the reaction equation Zn ( s ) + 2 HCl ( aq ) ⟶ ZnCl 2 ( aq ) + H 2 ( g ) How many milliliters of 5.00 M HCl ( aq ) are required to react with 3.15 g Zn ( s ) ?

We need 19.3 mL of HCl

Explanation:

Step 1: Data given

Molarity HCl = 5.00 M

Mass Zn = 3.15 grams

Step 2: The balanced equation

Zn (s) + 2HCl (aq) ⟶ ZnCl2 (aq) + H2 (g)

Step 3: Calculate moles Zn

Moles Zn = mass Zn / molar mass Zn

Moles Zn = 3.15 grams  / 65.38 g/mol

Moles Zn = 0.0482 moles

Step 4: Calculate moles HCl needed

For 2 moles HCl we need 1 mol Zn to produce 1 mol ZnCl2 and 1 mol H2

For 0.0482 moles Zn we need 2*0.0482 = 0.0964 moles HCl needed

Step 5: Calculate volume needed

Molarity = moles / volumes

Volumes = moles / molarity

Volume HCl needed = 0.0964 moles HCl /5.00 M

Volume HCl needed = 0.01928 L = 19.3 mL

We need 19.3 mL of HCl

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year