Question

If 0.613 g of magnesium hydroxide reacts with 0.960 g of sulfuric acid, what is the mass of magnesium sulfate produced? Mg(OH)2(s)+H2SO4(l)→MgSO4(s)+H2O(l) Express your answer with the appropriate units.

Answer 1

Molar mass of Mg(OH)2,

MM = 1*MM(Mg) + 2*MM(O) + 2*MM(H)

= 1*24.31 + 2*16.0 + 2*1.008

= 58.326 g/mol

mass(Mg(OH)2)= 0.613 g

use:

number of mol of Mg(OH)2,

n = mass of Mg(OH)2/molar mass of Mg(OH)2

=(0.613 g)/(58.33 g/mol)

= 1.051*10^-2 mol

Molar mass of H2SO4,

MM = 2*MM(H) + 1*MM(S) + 4*MM(O)

= 2*1.008 + 1*32.07 + 4*16.0

= 98.086 g/mol

mass(H2SO4)= 0.96 g

use:

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(0.96 g)/(98.09 g/mol)

= 9.787*10^-3 mol

Balanced chemical equation is:

Mg(OH)2 + H2SO4 ---> MgSO4 + 2 H2O

1 mol of Mg(OH)2 reacts with 1 mol of H2SO4

for 1.051*10^-2 mol of Mg(OH)2, 1.051*10^-2 mol of H2SO4 is required

But we have 9.787*10^-3 mol of H2SO4

so, H2SO4 is limiting reagent

we will use H2SO4 in further calculation

Molar mass of MgSO4,

MM = 1*MM(Mg) + 1*MM(S) + 4*MM(O)

= 1*24.31 + 1*32.07 + 4*16.0

= 120.38 g/mol

According to balanced equation

mol of MgSO4 formed = (1/1)* moles of H2SO4

= (1/1)*9.787*10^-3

= 9.787*10^-3 mol

use:

mass of MgSO4 = number of mol * molar mass

= 9.787*10^-3*1.204*10^2

= 1.178 g

Answer: 1.18 g