Question

How many grams are in 9.97 moles of Be(NO3)2?
Use two digits past the decimal for all values.

Answer 1

1,869.97 grams of Be(NO3)2

Step-by-step explanation:

Be(NO3)2 = Be N2 O6

Be=9.012182g/mole

N2=28.0134g/mole

O6=96g/mole

therefore Be(NO3)2 gives you 187.56g in one mole

so 9.97 moles means there is 9.97 times more

9.97mole Be(NO3)2 * 187.56g Be(NO3)2/1mole Be(NO3)2 = 1,869.97g of Be(NO3)2