What is the number of moles of calcium cyanide in 320 g calcium cyanide?

Question

Luna Chen · Accepted Answer

As always, #"Number of moles"-="Mass"/"Molar mass"=???mol#

Number of moles of #Ca(C-=N)_2=(320*g)/(92.11*g*mol^-1)~=3*mol#.

The equation is dimensionally consistent in that #1/(mol^-1)=1/(1/(mol))=mol#, as required for an answer in #"moles."#

This material is used extensively (and very destructively) in the gold mining industry. Solutions of the salt can leach the gold and silver content out of ores.......I would not like to be downwind of such an operation.

Xavier Avery · Answer

undefined To find the number of moles of calcium cyanide, you divide the given mass of calcium cyanide by its molar mass.

The molar mass of calcium cyanide (Ca(CN)₂) is:
$$ \text{Molar mass of Ca(CN)_2} = \text{Atomic mass of Ca} + 2 \times \text{Atomic mass of C} + 2 \times \text{Atomic mass of N} $$

$$ = 40.08 \text{ g/mol} + 2 \times 12.01 \text{ g/mol} + 2 \times 14.01 \text{ g/mol} $$

$$ = 40.08 \text{ g/mol} + 24.02 \text{ g/mol} + 28.02 \text{ g/mol} $$

$$ = 92.11 \text{ g/mol} $$

Now, calculate the number of moles:
$$ \text{Number of moles} = \frac{\text{Mass of calcium cyanide}}{\text{Molar mass of calcium cyanide}} $$

$$ = \frac{320 \text{ g}}{92.11 \text{ g/mol}} $$

$$ ≈ 3.47 \text{ mol} $$

Therefore, there are approximately 3.47 moles of calcium cyanide in 320 grams of calcium cyanide.