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# Answer to Question #182057 in General Chemistry for Lystyn White

Question #182057

### 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?

1
2021-04-16T04:59:03-0400

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"

2)

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

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"

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"

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