Question

I need help with number 8 please help ASAP.

Answer 1

Answer:

33.33% = 33%

Explanation:

MgCO3(s) + 2HCl (aq) --> MgCl2(aq) + H20(l) + CO2(g)

1 mole of MCO3 will produce → 1 mole of CO2

We need to get the number of mole of CO2:

and when we have 0.22 g of CO2, so number of mole = mass / molar mass

Moles = 0.22 g / 44 g/mol = 0.005 mole

Moles of Mg = moles of CO2 = 0.005 mole

Mass of Mg = moles * molar mass

= 0.005 * 84 /mol = 0.42 g

Percent of MgCO3 by mass of Mg = 0.42 g / 1.26 * 100

=33.33 %