#### Answer

(a) $CaCO_3 + 2HCl -- \gt CaCl_2 + CO_2 + H_2O$
(b) $ 4.52 \times 10^{23} atoms$
(c) $9.03 \times 10^{22}molecules(CO_2)$

#### Work Step by Step

(a)
1. Begin by balancing the elements that appear only once in each side:
- Balance the $Cl$:
The subscript for chlorine in the products is "2", from $CaCl_2$. Therefore, we should put a "2" as the coefficient of $HCl$
$CaCO_3 + 2HCl -- \gt CaCl_2 + CO_2 + H_2O$
- The equation is balanced.
(b) Each calcium carbonate molecules $CaCO_3$ has: 1 Calcium, 1 carbon and 3 oxygens, which gives us a total of 5 atoms. Use that as a conversion factor:
$0.150mol(CaCO_3) \times \frac{5atoms}{1(CaCO_3)} = 0.750mol-atoms$
Convert that number to atoms: (1 mol = $6.022 \times 10^{23}$)
$0.750 mol-atoms \times \frac{6.022 \times 10^{23}}{1 mol} = 4.52 \times 10^{23} atoms$
(c)
According to the balanced equation, each mole of calcium carbonate produces 1 mole of carbon dioxide ($CO_2$). Use that as a conversion factor:
$0.150mol(CaCO_3) \times \frac{1mol(CO_2)}{1mol(CaCO_3)} = 0.150mol(CO_2)$
Convert that number to molecules: (1 mol = $6.022 \times 10^{23}molecules$)
$0.150mol(CO_2) \times \frac{6.022 \times 10^{23} molecules}{1mol} = 9.03 \times 10^{22}molecules(CO_2)$