Question

ce cantitate de substanta ce cupru va interactiona 105g de solutie hno3 cu niu=60%?calculati cantitatea de substanta a sarii si v gaz (cu) c rezulta din reactie

Answer 1

ms = 105g HNO₃
C = 60%
n Cu = ? moli , m Cu = ? (g)
n sare = ? (moli), m sare = ? (g)
V gaz  = ? (L)

C = md/ms*100 => md = C*ms/100 = 60*105/100 = 63g HNO₃

MHNO₃ = 63g/mol ,   n = m/M = 63/63 = 1 mol HNO₃

3Cu + 8HNO₃ ⇒ 3Cu(NO₃)₂ + 2NO↑ + 4H₂O

3 moli Cu...............8 moli HNO₃
x moli Cu...............1 mol HNO₃

x = 3/8 = 0,375 moli Cu

MCu = 64g/mol,   n = m/M ⇒ m = n*M = 0,375*64 = 24g Cu

8 moli HNO₃.....................3 moli Cu(NO₃)₂
1 mol HNO₃....................x moli Cu(NO₃)₂

x = 3*1/8 = 0,375 moli Cu(NO₃)₂

MCu(NO₃)₂ = ACu + 2(AN+3*AO) = 64+2(14+3*16) = 188g/mol

n = m/M ⇒ m = n*M = 0,375*188 = 70,5g Cu(NO₃)₂

8 moli HNO₃.........................2*22,4 L NO
1 mol HNO₃..................................x L NO

x = 44,8/8 = 5,6 L NO