If 1.0L of spilled 1M HCl needed to be neutralized using solid sodium carbonate, how many grams of solid would be required?
a)53g
b)106g
c)1060g
d)530g
53
To determine the number of grams of solid sodium carbonate required to neutralize the spilled 1M HCl, we need to use the concept of stoichiometry.
The balanced chemical equation for the reaction between sodium carbonate (Na2CO3) and hydrochloric acid (HCl) is as follows:
Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O
From the balanced equation, we can see that one mole of sodium carbonate reacts with two moles of hydrochloric acid.
Given that the spilled HCl has a concentration of 1M and a volume of 1.0L, it means that there is one mole of HCl present.
To neutralize the HCl, we need an equal number of moles of sodium carbonate.
Therefore, we require the same number of moles of Na2CO3 as the number of moles of HCl, which is 1 mole.
The molar mass of Na2CO3 is 106 g/mol, which means that 1 mole of Na2CO3 weighs 106 grams.
Thus, the correct answer is b) 106g, as we would require 106 grams of solid sodium carbonate to neutralize the spilled 1M HCl.
To determine how many grams of solid sodium carbonate would be required to neutralize the spilled 1.0L of 1M HCl, we can use the concept of stoichiometry and balanced chemical equations.
The balanced chemical equation for the reaction between sodium carbonate (Na2CO3) and hydrochloric acid (HCl) is:
2 Na2CO3 + HCl -> 2 NaCl + H2O + CO2
From the equation, we can see that 2 moles of sodium carbonate react with 1 mole of hydrochloric acid. Since the concentration of the HCl is given as 1M (1 mole/L), we can infer that there is 1 mole of HCl in the spilled 1.0L.
To calculate how many grams of sodium carbonate are required, we need to use the molar mass of Na2CO3, which is 106 g/mol.
Now, we can set up the following conversion:
1 mole of HCl * (2 moles of Na2CO3 / 1 mole of HCl) * (106 g of Na2CO3 / 1 mole of Na2CO3)
By canceling units, we find that:
1 mole of HCl * 2 * 106 g of Na2CO3 / 1 mole of HCl = 212 g of Na2CO3
Therefore, approximately 212 grams of solid sodium carbonate would be required to neutralize the spilled 1.0L of 1M HCl.
Looking at the given options, none of them match exactly. However, the closest option is (c) 1060g. It seems that the options provided may have a typo, and the correct answer should be 212g, not 106g or 1060g.
Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
How many moles HCl do you have to neutralize? That will be moles = M x L = ?
How many moles Na2CO3 will that take. It will take half as many moles as HCl.
How many grams is that? moles = grams/molar mass.