The answer is 33.33 %
The explanation:
According to the reaction equation:
MgCO3(s) + 2HCl (aq) --> MgCl2(aq) + H20(l) + CO2(g)
we can see that 1 mole of MCO3 will produce → 1 mole of CO2
-Now we need o get 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 %
Answer: There are 
molecules present in 6.71 moles of 
.
Explanation:
Given: Moles of = 6.71 mol
According to the mole concept, 1 mole of every substance contains 
molecules.
Therefore, molecules present in 6.71 moles are calculated as follows.
Thus, we can conclude that there are molecules present in 6.71 moles of .
<span>NaCl is one and that is the only one that I know is for sure.
i hope this help!!!!!!!!!!</span>
