What mass of MgCl2 will be produced if 8.29 g of Mg(OH)2 reacts? Answer in units of g.

Question

part 1

Answer for part 1 and part2 in units of g Milk of magnesia, a suspension of Mg(OH)2 in water, reacts with stomach acid (HCl) in a neutralization reaction.

Mg(OH)2(s) + 2 HCl(aq) −→ 2 H2O(ℓ) + MgCl2(aq)

What mass of MgCl2 will be produced if 8.29 g of Mg(OH)2 reacts? Answer in units of g.

part 2

What mass of HCl is required to completely react with 8.29 g of Mg(OH)2? Answer in units of g.

Answer 1

First you find the moles of Mg(OH)₂ . You do this by using the mass of Mg(OH)₂ dividing it by the RMM (58.305gmol-1)

Mg(OH)₂ moles=8.29g/58.305gmol-1=0.142moles

Because the mole ratio of Mg(OH)₂ to MgCl₂ is 1:1 , you use this to get the amount of moles of MgCl₂

Mg(OH)₂+ 2HCl----->MgCl_2(aq) +2H₂O_(l)

You have 0.142moles of MgCl₂ .To0 get the mass you multiply moles by RMM.(94 gmol-1)

0.142 moles*94gmol-1=13.348g of MgCl₂

The mole ratio of Mg(OH)₂ to HCl is 1:2 so 0.142 moles of Mg(OH)₂ will need 0,282 moles of HCl

The mass of HCL would be 0.282moles*36gmol-1=10.224g

Answer 2

Find an answer here.

https://youtu.be/45mKzpJ5iIg