How many moles of sodium cyanide (NaCN) would be needed to produce 4.2 moles of sodium sulfate (Na2SO4)? (4 points)
H2SO4 + 2NaCN → 2HCN + Na2SO4
O 2.1 mol NaCN
O 4.2 mol NaCN
O 8.4 mol NaCN
O 12.0 mol NaCN

c

Explanation:

8.4

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.

it should be all of them because it asks if you included all of those in your answer

explanation: