Ammonium chloride and sodium nitrite react to form sodium chloride, water and nitrogen gas. If 6.33 moles of ammonium chloride react with sodium nitrite, how many grams of sodium chloride will be produced?
NH4Cl + NaNO2 ==> NaCl + N2 + 2H2O
6.33 mols NH4Cl x (1 mol NaCl/1 mol NH4Cl) = 6.33 mols NaCl formed.
Then g = mols x molar mass.
To determine the grams of sodium chloride produced, we need to use the balanced chemical equation and mole-to-mole ratios.
The balanced equation for the reaction is:
NH4Cl + NaNO2 -> NaCl + H2O + N2
From the equation, we can see that 1 mole of NH4Cl reacts with 1 mole of NaCl. Therefore, the moles of NaCl produced will be the same as the moles of NH4Cl used.
Given that 6.33 moles of NH4Cl react with NaNO2, we can conclude that 6.33 moles of NaCl will be produced.
To convert moles to grams, we need to know the molar mass of NaCl, which is 58.44 g/mol.
So, the mass of NaCl produced can be calculated by multiplying the number of moles of NaCl by its molar mass:
Mass of NaCl = 6.33 moles * 58.44 g/mol
Calculating this:
Mass of NaCl = 370.21 g
Therefore, approximately 370.21 grams of sodium chloride will be produced when 6.33 moles of ammonium chloride react with sodium nitrite.