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nexus9112 [7]
3 years ago
11

How many mL of stomach acid (assume it is .035 M HCI) can be neutralized by an antacid tablet containing .231g

Chemistry
1 answer:
ASHA 777 [7]3 years ago
4 0
Mg(OH)2(aq.) + 2HCl(aq.) --> MgCl2(aq.) + 2H2O(l) 

<span>n(HCl)/n(Mg(OH)2)= 2/1 </span>

<span>n(HCl)= 2*n(Mg(OH)2) </span>

<span>c= n/V; n= c*V </span>

<span>n= m/M </span>

<span>c(HCl)*V(HCl)= 2*m(Mg(OH)2)/M(Mg(OH)2) </span>

<span>V(HCl)= 2*m(Mg(OH)2)/(M(Mg(OH)2)*c(HCl))= 2 * 0,231 g / (58,326 g mole^-1 * 0,035 mole dm^-3)= 0,226 dm^3= 226 cm^3 (cm^3= mL) </span>
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