I have to write the equation for the following and I am not figuring it out.I was out with flu and so lost...please help.
How many moles of Hydrochloric acid are needed to dissolve 39.0 grams of aluminum Hydroxide?
How many grams of precipitate will form a reaction between 10g of calcium bromide and an excess of silvernitrate?
How many grams of gas will evolve in reaction between 20.0g of chalk (CaCO3) and an excess of sulfiric acid?
The middle one.
CaBr2 + 2AgNO3 --> 2AgBr + Ca(NO3)2
mols CaBr2 = g/molar mass
Using the coefficients in the balanced equation, convert mols CaBr2 to mols AgNO3.
Now convert mols AgNO3 to g. g = mols x molar mass.
To answer these questions, we need to use the concept of stoichiometry. Stoichiometry is a way of calculating the amounts of reactants and products involved in a chemical reaction.
1. How many moles of Hydrochloric acid are needed to dissolve 39.0 grams of aluminum hydroxide?
First, we need to determine the molar mass of aluminum hydroxide. Aluminum has a molar mass of 27 g/mol, and hydroxide has a molar mass of 17 g/mol. Add these together to get the molar mass of aluminum hydroxide, which is 39 g/mol.
Next, we can use the molar mass to convert grams of aluminum hydroxide to moles. Divide the given mass (39.0 g) by the molar mass (39 g/mol) to obtain the number of moles of aluminum hydroxide.
Now, we need to look at the balanced equation for the reaction between hydrochloric acid and aluminum hydroxide. The balanced equation is:
2 Al(OH)3 + 6 HCl -> 2 AlCl3 + 6 H2O
From the balanced equation, we can see that 2 moles of aluminum hydroxide react with 6 moles of hydrochloric acid. Therefore, the number of moles of hydrochloric acid needed will be three times the number of moles of aluminum hydroxide.
Multiply the number of moles of aluminum hydroxide by 3 to get the number of moles of hydrochloric acid required.
2. How many grams of precipitate will form in a reaction between 10g of calcium bromide and an excess of silver nitrate?
To answer this question, we need to know the balanced equation for the reaction between calcium bromide and silver nitrate. Let's assume the balanced equation is:
CaBr2 + 2AgNO3 -> 2AgBr + Ca(NO3)2
From the balanced equation, we can see that 1 mole of calcium bromide reacts with 2 moles of silver nitrate to form 2 moles of silver bromide.
First, calculate the number of moles of calcium bromide by dividing the given mass (10g) by the molar mass of calcium bromide (which you can find by adding the molar masses of calcium and bromine together).
Next, using the stoichiometric ratio from the balanced equation, multiply the number of moles of calcium bromide by the stoichiometric ratio of silver bromide to calcium bromide (2 moles of AgBr / 1 mole of CaBr2). This will give you the number of moles of silver bromide.
Finally, convert the moles of silver bromide to grams by multiplying by the molar mass of silver bromide.
3. How many grams of gas will evolve in the reaction between 20.0g of chalk (CaCO3) and an excess of sulfuric acid?
First, calculate the number of moles of chalk (CaCO3) by dividing the given mass (20.0g) by the molar mass of chalk (which you can find by adding the molar masses of calcium, carbon, and oxygen together).
Next, the balanced equation for the reaction between calcium carbonate and sulfuric acid is:
CaCO3 + H2SO4 -> CaSO4 + CO2 + H2O
From the balanced equation, we can see that 1 mole of calcium carbonate reacts to produce 1 mole of carbon dioxide.
Using the stoichiometric ratio, multiply the number of moles of calcium carbonate by the stoichiometric ratio of carbon dioxide to calcium carbonate (1 mole of CO2 / 1 mole of CaCO3).
Finally, convert the moles of carbon dioxide to grams by multiplying by the molar mass of carbon dioxide.