Question

How many grams of magnesium oxide remain after 18.50 g of magnesium oxide reacts with 30.30 g hydrochloric acid to produce magnesium chloride and water?

Please show clear work on paper, thanks!

Answer 1

Molar mass of MgO,

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

= 1*24.31 + 1*16.0

= 40.31 g/mol

mass(MgO)= 18.5 g

number of mol of MgO,

n = mass of MgO/molar mass of MgO

=(18.5 g)/(40.31 g/mol)

= 0.4589 mol

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass(HCl)= 30.3 g

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(30.3 g)/(36.458 g/mol)

= 0.8311 mol

Balanced chemical equation is:

MgO + 2 HCl ---> MgCl2 + 2 H2O

1 mol of MgO reacts with 2 mol of HCl

for 0.4589 mol of MgO, 0.9179 mol of HCl is required

But we have 0.8311 mol of HCl

so, HCl is limiting reagent

we will use HCl in further calculation

According to balanced equation

mol of MgO reacted = (1/2)* moles of HCl

= (1/2)*0.8311

= 0.4155 mol

mol of MgO remaining = mol initially present - mol reacted

mol of MgO remaining = 0.4589 - 0.4155

mol of MgO remaining = 0.0434 mol

mass of MgO remaining,

m = number of mol * molar mass

= 4.34*10^-2 mol * 40.31 g/mol

= 1.75 g

Answer: 1.75 g