<span>CH4 + 4 Cl2 â†’ CCl4 + 4 HCl (4.00 mol CH4) x (1/1) x (0.70) = 2.80 mol CCl4 (4.00 mol CH4) x (4/1) x (0.70) = 11.2 mol HCl CCl4 + 2 HF â†’ CCl2F2 + 2 HCl (2.80 mol CCl4) x (2/1) x (0.70) = 3.92 mol HCl 11.2 mol + 3.92 mol = 15.1 mol HCl from both steps</span>

8 0

3 years ago

What is the answer please tell me

Nastasia [14]

Answer:

you need to include the bottom portion, not enough info

Explanation:

5 0

2 years ago

a 10.99g sample of NaBr contains 22.34% Na by mass. Considering the law of constant composition (define proportions), how many g

leonid [27]

Given :

A 10.99 g sample of NaBr contains 22.34% Na by mass.

To Find :

How many grams of sodium does a 9.77g sample of sodium bromine contain.

Solution :

By law of constant composition , in any given chemical compound, the elements always combine in the same proportion with each other.

Therefore , percentage of Na by mass in NaBr will be same for every amount .

Percentage of Na in 9.77 g NaBr is 22.34 % too .

Gram of Na = $9.77\times \dfrac{22.34}{100}=2.18\ g$ .

Hence , this is the required solution .

7 0

2 years ago