1- Qual massa de ácido sulfúrico (H 2 SO 4 ) será necessária para preparar 2 litros de uma solução na concentração de 3 mol/L?
Dado: M(H 2 SO 4 )= 98 g/mol. M= 2.1+32+4.16= 98g
Dados do exercício:
m 1 = ? (é o que se quer encontrar)
M 1 = 98 g/mol
V (L) = 2 L
C = 3 mol/L

2- (Fuvest-SP) A concentração de íons fluoreto em uma água de uso
doméstico é de 5,0 · 10 –5 mol/litro. Se uma pessoa tomar 3,0 litros
dessa água por dia, ao fim de um dia, a massa de fluoreto, em
miligramas, que essa pessoa ingeriu é igual a: (massa molar do
fluoreto: 19,0 g/mol)

Dados do exercício:
m 1 = ? (é o que se quer encontrar)
M= 19,0 g/mol
V (L) = 3 L
C= 5 . 10 -5 mol/L
Aplicando os dados na fórmula da concentração em mol/L,
temos:

a chemical reaction

explanation:

elements make up everything, and they combine to make compounds.

the answer is c

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2