Find the limiting reagent.
25.7g of NH3*(1 mole/17.031 g)= moles of NH3
22.4g of Na *(1 mole/22.990 g)= moles of Na
Since moles of Na= moles of NH3=moles of NaNH2
The lowest number of moles from the above calculations multiplied by molecular weight will give you the amount of NaNH2
moles of limiting reagent*(39.01 g/mol)= mass of NaNH2
2NH3+ 2Na---) 2NaNH2+1H2
assuming that you start with 25.7 g of ammonia gas and 22.4 g of sodium metal and asuming that the reaction goes to completion what is the mass in grams of NaNH2
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