Question

2A + 3B--> A2B3 If 3mL of 0.04M A, 4mL of 0.02M B and 3mL of H2O was into a reaction test tube, what are the concentrations of A and B in the test tube.

Answer 1

`[A]=0.012M`

`[B]=0.008M`

Step-by-step explanation:

Hello,

In this case, given the concentration and volume of each reactant, we first compute their moles in the solution:

`n_A=0.003L*0.04mol/L=1.2x10^(-4)mol\\\\n_B=0.004L*0.02mol/L=8x10^(-5)mol`

Thus, since the total volume is computed given the volume of each reactant as well as the extra water, we have:

`V=0.003L+0.004L+0.003L=0.010L`

Thus, the concentrations of A and B are:

`[A]=(1.2x10^(-4)mol)/(0.010L)=0.012M`

`[B]=(8x10^(-5)mol)/(0.010L)=0.008M`

Best regards.