Question

4 Zn + 10 HNO₃     -->    4 Zn(NO₃)₂  + N₂O   + 5 H₂O

How many g of HNO₃ would be needed to produce 34.6 L N₂O?

How many mol N₂O would be produced from 134.5 g Zn reacting with excess HNO₃?If 125 g H₂O were produced, how many g Zn were used?

How many g Zn would be needed to produce 2.55 x 10²⁴ molecules N₂O?

Answer 1

1)

Assuming STP conditions,

$m(HNO_3)=34.6L(N_2O)\times\frac{1mol(N_2O)}{22.4L(N_2O)}\times\frac{10mol(HNO_3)}{1mol(N_2O)}\times\frac{63.01g(HNO_3)}{1mol(HNO_3)}=973g$

Answer: 973 g

2)

$n(N_2O)=134.5g(Zn)\times\frac{1mol(Zn)}{65.38g(Zn)}\times\frac{1mol(N_2O)}{4mol(Zn)}=0.5143mol$

Answer: 0.5143 mol

3)

$m(Zn)=125g(H_2O)\times\frac{1mol(H_2O)}{18.015g(H_2O)}\times\frac{4mol(Zn)}{5mol(H_2O)}\times\frac{65.38g(Zn)}{1mol(Zn)}=363g$

Answer: 363 g

4)

$m(Zn)=2.55\cdot10^{24}(molecules\;N_2O)\times\frac{1mol(N_2O)}{6.022\cdot10^{23}(molecules\;N_2O)}\times\frac{4mol(Zn)}{1mol(N_2O)}\times\frac{65.38g(Zn)}{1mol(Zn)}=1110g$

Answer: 1110 g