Assuming stomach acid is 0.1 m hcl, how many grams of stomach acid would have been neutralized by your tablet (use the average value)?

HCl = 0.1 M. 

find grams of Acid to neutralize the tablet. 

n = 0.00739 mol of CO3-2. 

D = 1.02 g/ml. 

Find the moles of CO3-2.

CO3-2 + 2HCl <---> H2CO3 + 2Cl-.

therefore,

we need 2 mol of CO3-2 to neutralize 1 mol of HCl.

since we have 0.00739 mol of CO3-2, that will neutralize 2X mol of HCl.

2*0.00739 = 0.01478 mol of HCl will be neutralized.

Now, find the mass of Acid :

M = mol/V,

V = mol/M.

Substitute data of acid :

V = 0.01478 mol / 0.1 M = 0.1478 Liters of stomach acid

or 147.8 ml of stomach acid

now find mass:

D = mass/V.

mass = D*V = 1.02*147.8 = 150.76 grams of Stomach Acid will be neutralized

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

the answer is b. dilution