Question

For each of the following reactions calculate the mass (in grams) of both the reactants that are required to form 15.39g of the following products.

a. 2K(s) + Cl2(g) → 2Cl(aq)
b. 4Cr(s) + 302(g) → 2Cr2O3(s)
c. 35r(s) + N2(g) → SraNa(s)

Answer 1

a.

`m_K=8.056gK\\ \\m_(Cl_2)=4.028gCl_2`

b.

`m_(Cr)=10.51gCr\\ \\m_(O_2)=4.851gO_2`

c.

`m_(Sr)=13.88gSr\\\\m_(N_2)=1.479gN_2`

Step-by-step explanation:

Hello,

In this case, we proceed via stoichiometry in order to compute the masses of all the reactants as shown below:

a.
`2K+Cl_2\rightarrow 2KCl`

`m_K=15.36gKCl*(1molKCl)/(74.55gKCl)*(2molK)/(2molKCl)* (39.1gK)/(1molK)=8.056gK\\ \\m_(Cl_2)=15.36gKCl*(1molKCl)/(74.55gKCl)*(1molCl_2)/(2molKCl)* (70.9gCl_2)/(1molCl_2)=4.028gCl_2`

b.
`4Cr+ 3O_2\rightarrow 2Cr_2O_3`

`m_(Cr)=15.36gCr_2O_3*(1molCr_2O_3)/(152gCr_2O_3l)*(4molCr)/(2molCr_2O_3)* (52gCr)/(1molCr_2O_3)=10.51gCr\\ \\m_(O_2)=15.36gCr_2O_3*(1molCr_2O_3)/(152gCr_2O_3l)*(3molO_2)/(2molCr_2O_3)* (32gO_2)/(1molCr_2O_3)=4.851gO_2`

c.
`3Sr(s) + N_2(g) \rightarrow Sr_3N_2`

`m_(Sr)=15.36gSr_3N_2*(1molSr_3N_2)/(290.86gSr_3N_2)*(3molSr)/(1molSr_3N_2)* (87.62gSr)/(1molSr)=13.88gSr\\\\m_(N_2)=15.36gSr_3N_2*(1molSr_3N_2)/(290.86gSr_3N_2)*(1molN_2)/(1molSr_3N_2)* (28gN_2)/(1molN_2)=1.479gN_2`

Regards.