How many grams of NH3 are needed when 2.96 g of Cr(NO3)3 react Cr(NO3)3 + 6NH3 → Cr(NH3)6(NO3)3

Question

How many grams of NH₃ are needed when 2.96 g of Cr(NO₃)₃ react according to the following balanced equation?

Cr(NO₃)₃ + 6NH₃ → Cr(NH₃)₆(NO₃)₃

Answer 1

This is a perfect example of how to utilize stoichiometry to convert between grams and moles of a substance. When asked to convert anything in chemistry, it is usually safe to assume stoichiometry is the easiest way to do so. It is important to note that the typical order for this kind of conversion is to go from Grams - Moles - Moles - Grams.

Step 1: Convert grams of Cr(NO₃)₃ to moles by using its mass

2.96g Cr(NO₃)₃ x 1 mol Cr(NO₃)₃

238.011g Cr(NO₃)₃

Now that we are in moles of Cr(NO₃)₃ It is time to continue the equation

Step 2: Use the balanced equation to convert moles of Cr(NO₃)₃ to moles of NH₃

x 1 mol Cr(NO₃)₃ x 6 mol NH₃

238.011g Cr(NO₃)₃ 1 mol Cr(NO₃)₃

Now that we are in moles of NH₃ we go to grams of NH₃

Step 3: Use the mass of NH₃ to covert to grams

x 6 mol NH₃ x 17.031g NH₃

1 mol Cr(NO₃)₃ 1 mol NH₃

Step 4: Now that we have finished doing the conversions, once you cancel out terms you should be left in grams of NH₃

2.96g Cr(NO₃)₃ x 1 mol Cr(NO₃)₃ x 6 mol NH₃ x 17.031g NH₃

1 238.011g Cr(NO₃)₃ 1 mol Cr(NO₃)₃ 1 mol NH₃

Step 5: Now that everything is canceled out it is time for the last calculation.

2.96 x 6 x 17.031g NH₃ = 1.271g NH₃

238.011

ALL DONE!