Calcium carbonate reacts with aqueous hcl

Calcium carbonate reacts with aqueous $HCl$ to gives $CaC$ which is required to react completely with $25ml$ of $0.75M$$HCl$ is:$\\left( A \\right)$ $1.825g$$\\left( B \\right)$ $0.9375g$$\\left( C \\right)$ $1.8357g$$\\left( D \\right)$ $0.46875g$ Calcium carbonate reacts with aqueous $HCl$ to gives $CaC$ which is required to react completely with $25ml$ of $0.75M$$HCl$ is: $\left( A \right)$ $1.825g$ $\left( B \right)$ $0.9375g$ $\left( C \right)$ $1.8357g$ $\left( D \right)$ $0.46875g$ Hint: First we have to know the molar mass of a chemical compound is defined as the mass of a sample of that compound divided by the amount of substance in that sample, measured in moles. The molar mass is a bulk, not molecular, property of a substance. The molar mass of a compound is calculated by adding there standard atomic masses in \[\left( g$ .

Calcium carbonate reacts with aqueous HCl to give CaCl 2 and CO 2 according to the reaction given below CaCO 3 ( s ) Calcium Carbonate + 2 HCl ( aq ) Hydrochloric acid ⟶ CaCl 2 ( aq ) Calcium Chloride + CO 2 ( g ) Carbon dioxide + H 2 O ( l ) Water What mass of CaCO 3 is required to react completely with 25 mL of 0.75 M HCl? 1) 0.6844 2) 0.9375 3) 0.4265 4) 0.2785 The given equation is CaCO 3 ( s ) Calcium Carbonate + 2 HCl ( aq ) Hydrochloric acid ⟶ CaCl 2 ( aq ) Calcium Chloride + CO 2 ( g ) Carbon dioxide + H 2 O ( l ) Water Number of moles of HCl = Molarity x Volume = 0 . 75 M × 0 . 025 L = 0 . 01875 mol As, the mole ratio between CaCO 3 and HCl is 1:2. So, the number of moles of CaCO 3 should be half of the number of moles of HCl Number of moles of CaCO 3 = 1 2 × 0 . 01875 mol = 0 . 009375 mol Molar mass of CaCO 3 = 100 g / mol Mass of Calcium Carbonate = Moles ×Molar mass = 0 . 009375 mol × 100 g / mol = 0 . 9375 g So option (2) is correct Q. Calcium carbonate reacts with aqueous HCl to give C a C l 2 and C O 2 according to the reaction given below C a C O 3 ( s ) + 2 H C l ( a q ) ⟶ C a C l 2 ( a q ) + C O 2 ( g ) + H 2 O ( l ) What mass of C a C l 2 will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of C a C O 3 ? Name the limiting reagent. Calculate the number of moles of C a C l 2 formed in the reaction.

Calcium carbonate reacts with aqueous HCl to give CaCl 2 and CO 2 according to the reaction given below: CaCO 3 (s) + 2HCl (aq) → CaCl 2(aq) + CO 2(g) + H 2O(l) What mass of CaCl 2 will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of CaCO3? Name the limiting reagent. Calculate the number of moles of CaCl 2 formed in the reaction. According to given equation 1 mol of CaCO 3 (s) requires 2 mol of HCl (aq). Hence, for the reaction of 10 mol of CaCO 3 (s) number of moles of HCl required would be: But we have only 0.19 mol HCl (aq), hence, HCl (aq) is limiting reagent. So amount of CaCl 2 formed will depend on the amount of HCl available. Since, 2 mol HCl (aq) forms 1 mol of CaCl 2, therefore, 0.19 mol of HCl (aq) would give:

1000 mL of.0.75 M HCl have 0.75 mol of HCl = 0.75 x 36.5 g = 24.375 g (Molar mass of HCl is 36.5) Mass of HCl in 25 mL of 0.75 M HCl = 24.375/1000 x 25 g = 0.6844 g According to the given chemical equation, CaCO 3(s) + 2HCl(ag) →CaCl 2(aq) + CO 2(g) + H 2O(Z) 2 mol of HCl i.e. 73 g HCl react completely with 1 mol of CaCO 3g i.e. 100 g .’. 0.6844 g HCl reacts completely with CaCO 3= 100/73 x 0.6844 g = 0.938 g