Calcium carbonate reacts with stomach acid (HCI) according to the following equation: Caco, (s) + 2 HCl(aq) - CaCl, (aq) + CO2(g) + H20 (1)
Tums, an antacid, contains CaCo3. If Tums is added to 25.7 mL of a 0.738 M HCl solution, how many grams of Co, are produced?

0.418 g of CO₂ will be produced

Explanation:

Equation of the reaction is given below:

CaCO₃ (s) + 2HCl (aq) > CaCl₂(aq) + CO₂ (g) + H₂O (l)

From the equation of reaction , 1 mole of CaCO₃ reacts with 2 molemolesmof HCl to produce 1 mole of CO₂

Number of moles of HCl in 25.7 ml of a 0.738 M solution of HCl is obtained using the formula below:

Number of moles = molarity × volume (Litres)

Number of moles of HCl = 0.738 M × 25.7 mL × 1 L/1000 mL = 0.0190 moles

Since the antacid, Tums will be excess, the limiting reactant is HCl

2 moles of HCl produces 1 mole of CO₂

0.0190 moles of HCl will produce 0.0190 moles ÷ 2 = 0.0095 molesof CO₂

Mass of 0.0095 moles of CO₂ = Numbe of moles × molar mass of CO₂

Molar mass of CO₂ = 44.0 g/mol

Mass of CO₂ produced = 0.00950 moles × 44.0 g/mol = 0.418 g of CO₂

Therefore, 0.418 g of CO₂ will be produced if Tums is added to 25.7 mL of a 0.738 M HCl solution

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong