Consider the reaction represented by the following unbalanced chemical equation NH3(g) + CL2(g) --> NH4CL(s)+NCL3(g)


#1

Consider the reaction represented by the following unbalanced chemical equation NH3(g) + CL2(g) --> NH4CL(s)+NCL3(g) What mass of NH4CL can be produced from 10.00g of NH3 and an excess of CL2

Answer:

First we must balance the equation :4NH3(g) + 2Cl2(g) --> 3NH4Cl(s)+NCl3(g)
Then we calculate number of moles in 10 g of NH3: 10 / (14.007 + 1.008 * 3) =
=0. 146791146 moles
Next we must determine the amount of moles of NH4Cl:
(3 * 0.146791146) / 4 = 0.440373437 moles NH4Cl
Lastly, we must convert moles to grams: 0.440373437(14.007+1.008*4 +35.453) =
= 23.55645589= 23.56 g