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Zinc and hydrochloric acid react to form hydrogen gas. In this reaction, zinc metal reacts with aqueous hydrochloric acid to form aqueous zinc chloride plus hydrogen gas:
Zn(s) + 2HCl(aq) --> ZnCl2(aq) + H2(g)
If you look carefully at the equation, oxygen is not found in any of the reactants or the products, so it would be impossible to get oxygen from this reaction.
What else can I help you with?
How many grams of zinc can react with 225ml of 0.200 m HCl solution?
Balanced equation first.Zn + 2HCl - ZnCl2 + H2now we find moles HCl by....Molarity = moles of solute/Liters of solution ( 225 ml = 0.225 Liters )0.200 M HCl = X moles/0.225 Liters= 0.045 moles HCl================Now, drive reaction backwards.0.045 moles HCl (1 mole Zn/2 mole HCl)(65.41 grams/1 mole Zn)= 1.47 grams zinc reacted----------------------------------
How do you prepare 0.5n HCL from 37 percent HCL?
To prepare 0.5N HCl from 37% HCl solution, you can use the formula C1V1 = C2V2 where C1 is the initial concentration, V1 is the initial volume, C2 is the desired concentration (0.5N), and V2 is the final volume. Calculate the volume of 37% HCl needed and dilute it to the desired volume with water.
How to prepare HCL solution?
To prepare a hydrochloric acid (HCl) solution, you can dilute concentrated hydrochloric acid with water. Always add the acid to water slowly while stirring, never add water to acid, as it can cause splattering. Be sure to wear appropriate protective gear and work in a well-ventilated area when handling HCl.
What volume of a 1.50 M HCl solution should you use to prepare 2.00 L of a 0.100 M HCl solution?
To prepare a 0.100 M HCl solution from a 1.50 M HCl solution, you need to use the dilution formula, which is M1V1 = M2V2. You would need to use (V_1 = \frac{M_2V_2}{M_1}) to calculate the volume needed. Plugging in the values, you would need to use ( V_1 = \frac{0.100 M \times 2.00 L}{1.50 M} = 0.133 L or 133 mL) of the 1.50 M HCl solution.
How do you prepare 0.2 mol per liter HCl from concentrated HCL?
To prepare 0.2 mol/L HCl from concentrated HCl (e.g., 37% HCl), you would need to dilute the concentrated HCl with water in the appropriate ratio. Since the concentrated HCl usually has a density of around 1.19 g/mL, you can use the formula M1V1 = M2V2 to calculate the volume of concentrated HCl needed. After calculating the volume of concentrated HCl required, add water to make up the final volume of 1 liter to achieve a 0.2 mol/L HCl solution.
Which acid and metal would you use to make zinc chloride?
Zinc (Zn) and hydrogen chloride HCl)
How many grams of zinc can react with 225ml of 0.200 m HCl solution?
Balanced equation first.Zn + 2HCl - ZnCl2 + H2now we find moles HCl by....Molarity = moles of solute/Liters of solution ( 225 ml = 0.225 Liters )0.200 M HCl = X moles/0.225 Liters= 0.045 moles HCl================Now, drive reaction backwards.0.045 moles HCl (1 mole Zn/2 mole HCl)(65.41 grams/1 mole Zn)= 1.47 grams zinc reacted----------------------------------
How do you prepare 0.5n HCL from 37 percent HCL?
To prepare 0.5N HCl from 37% HCl solution, you can use the formula C1V1 = C2V2 where C1 is the initial concentration, V1 is the initial volume, C2 is the desired concentration (0.5N), and V2 is the final volume. Calculate the volume of 37% HCl needed and dilute it to the desired volume with water.
What reagents would you use for the synthesis of 1-chlorobutane from 1-butanol?
anhydrous Zinc chloride and aq.HCl or CaCl2 and HCl.
How to prepare HCL solution?
To prepare a hydrochloric acid (HCl) solution, you can dilute concentrated hydrochloric acid with water. Always add the acid to water slowly while stirring, never add water to acid, as it can cause splattering. Be sure to wear appropriate protective gear and work in a well-ventilated area when handling HCl.
What volume of a 1.50 M HCl solution should you use to prepare 2.00 L of a 0.100 M HCl solution?
To prepare a 0.100 M HCl solution from a 1.50 M HCl solution, you need to use the dilution formula, which is M1V1 = M2V2. You would need to use (V_1 = \frac{M_2V_2}{M_1}) to calculate the volume needed. Plugging in the values, you would need to use ( V_1 = \frac{0.100 M \times 2.00 L}{1.50 M} = 0.133 L or 133 mL) of the 1.50 M HCl solution.
How do you prepare 0.2 mol per liter HCl from concentrated HCL?
To prepare 0.2 mol/L HCl from concentrated HCl (e.g., 37% HCl), you would need to dilute the concentrated HCl with water in the appropriate ratio. Since the concentrated HCl usually has a density of around 1.19 g/mL, you can use the formula M1V1 = M2V2 to calculate the volume of concentrated HCl needed. After calculating the volume of concentrated HCl required, add water to make up the final volume of 1 liter to achieve a 0.2 mol/L HCl solution.
How do you prepare zinc chloride powder from aqueous form?
To prepare zinc chloride powder from an aqueous solution, you can first evaporate the water by heating the solution. This will leave behind the solid zinc chloride. Make sure to perform this step in a well-ventilated area as zinc chloride can release fumes. Collect the dried zinc chloride crystals for use as a powder.
How to prepare 1m hcl from 35 percent hcl?
To prepare 1M HCl solution from 35% HCl solution, you would need to dilute the 35% HCl with water. Use the formula C1V1 = C2V2, where C1 is the initial concentration, V1 is the volume of the initial solution, C2 is the final concentration (1M), and V2 is the final volume (1 liter in this case). Calculate the volume of 35% HCl needed to achieve a 1M solution, then add water to make up the total volume to 1 liter.
How do you get zinc from zinc blende?
Zinc can be obtained from zinc blende (also known as sphalerite) through a process called roasting. First, the zinc blende is heated in the presence of oxygen to convert it into zinc oxide. Then, the zinc oxide is further heated with coke (a form of carbon) to reduce it to metallic zinc, which can be collected for further use.
How zinc react with acid?
Use any acid. HCl, HBr, HX. Zn + HX ==> ZnX + H2. The zinc forms a salt with the acid by getting oxidized. Hydrogen gas is released. With hydrochloric acid, the reaction is as follows: Zn + HCl ==> ZnCl2 + H2
How many grams of zinc chloride can be formed from the reaction of 7.96 g of zinc with excess HCl?
The equation for the reaction is Zn + 2 HCl -> ZnCl2. The gram atomic masses of zinc and chlorine are 65.39 and 35.453 respectively. Therefore, the number of grams of zinc chloride that can be formed from 7.96 g of zinc is: 7.96{[65.39 + 2(35.453)]/65.39} or about 16.6 g of zinc chloride, to the justified number of significant digits.
