Question

Решить в ионном виде 2HCL+CaCO3→CaCL+H2O+CO2↑ 2NaОH+H2SO4→Na2SO4+2H20 CU(OH)2↓+2HCL→CUCL2+2H2O

Answer 1

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Answer 2

Balancing Chemical Equations

To solve the given chemical equations in ionic form, we need to balance the equations by ensuring that the number of atoms of each element is the same on both sides of the equation. Let's solve each equation step by step.

1. 2HCl + CaCO3 → CaCl2 + H2O + CO2↑

To balance this equation, we need to ensure that the number of atoms of each element is the same on both sides. Here's how we can balance it:

2HCl + CaCO3 → CaCl2 + H2O + CO2↑

On the left side, we have 2 hydrogen atoms (H) and 2 chlorine atoms (Cl). On the right side, we have 2 hydrogen atoms (H) and 2 chlorine atoms (Cl). So, the hydrogen and chlorine atoms are already balanced.

Next, let's balance the carbon (C) and oxygen (O) atoms. On the left side, we have 1 carbon atom (C) and 3 oxygen atoms (O) from the carbonate ion (CO3^2-). On the right side, we have 1 carbon atom (C) and 1 oxygen atom (O) from the carbon dioxide (CO2) and 1 oxygen atom (O) from the water (H2O). To balance the carbon and oxygen atoms, we need to add a coefficient of 1 in front of CaCO3 and a coefficient of 1 in front of CO2:

2HCl + CaCO3 → CaCl2 + H2O + CO2

Now the equation is balanced.

2. 2NaOH + H2SO4 → Na2SO4 + 2H2O

To balance this equation, we need to ensure that the number of atoms of each element is the same on both sides. Here's how we can balance it:

2NaOH + H2SO4 → Na2SO4 + 2H2O

On the left side, we have 2 sodium atoms (Na), 2 oxygen atoms (O), and 2 hydrogen atoms (H). On the right side, we have 2 sodium atoms (Na), 1 sulfur atom (S), 4 oxygen atoms (O), and 4 hydrogen atoms (H). So, the sodium, oxygen, and hydrogen atoms are already balanced.

Next, let's balance the sulfur (S) atom. On the left side, we have 1 sulfur atom (S) from the sulfuric acid (H2SO4). On the right side, we have 1 sulfur atom (S) from the sodium sulfate (Na2SO4). To balance the sulfur atom, we need to add a coefficient of 1 in front of H2SO4 and a coefficient of 1 in front of Na2SO4:

2NaOH + H2SO4 → Na2SO4 + 2H2O

Now the equation is balanced.

3. Cu(OH)2 + 2HCl → CuCl2 + 2H2O

To balance this equation, we need to ensure that the number of atoms of each element is the same on both sides. Here's how we can balance it:

Cu(OH)2 + 2HCl → CuCl2 + 2H2O

On the left side, we have 1 copper atom (Cu), 2 hydrogen atoms (H), 2 chlorine atoms (Cl), and 4 oxygen atoms (O). On the right side, we have 1 copper atom (Cu), 2 chlorine atoms (Cl), and 4 hydrogen atoms (H). So, the copper, hydrogen, and chlorine atoms are already balanced.

Next, let's balance the oxygen (O) atoms. On the left side, we have 4 oxygen atoms (O) from the hydroxide ions (OH^-). On the right side, we have 4 oxygen atoms (O) from the water (H2O). To balance the oxygen atoms, we need to add a coefficient of 2 in front of H2O:

Cu(OH)2 + 2HCl → CuCl2 + 2H2O

Now the equation is balanced.

I hope this helps! Let me know if you have any further questions.