Question

in the reaction 4NH3(g)+5O2(g)-> 4NO(g)+ 6H2O(l)
a. How many grams of NH3 react with 73.4 g of O2?
b. How many grams of NO are produced from 73.4g of O2?

Answer 1

$4NH_3(g)+5O_2(g)\to4NO(g)+ 6H_2O(l)$

Moles of $O_2$ in $73.4\ g=\frac{73.4}{32}=2.29$

a. $5$ moles of $O_2$ reacts with $4$ moles of $NH_3$ .

$2.29$ moles of $O_2$ will react with $\frac{4}{5}\times 2.29=1.835$ moles$=1.835\times 17=31.195\ g$

b. $5$ moles of $O_2$ produce $4$ moles of $NO.$

$2.29$ moles of $O_2$ will produce $\frac{4}{5}\times 2.29=1.835$ moles$=1.835\times (14+16)\ g=55.05\ g$