Question

Please help and explain how to get the answer!

1. given the equation: 2Na+Cl2-->2NaCl
If 200 grams of NaCI is produced, how many grams of Na must be reacted with excess chlorine?
A. 58.43g Na
B. 78.65g Na
C. 22.98g Na
D. 3.4g Na

2. given the equation: 2K+2H2O-->2KOH+H2
if 23.5 grams of potassium are reacted with excess water, how many grams of potassium hydroxide will be formed?
A. 33.7g KOH
B. 56.08g KOH
C. 39.09g KOH
D. 17.99g KOH

Answer 1

Answer 1: The correct answer is option (B)

Answer 2: The correct answer is option (B)

Explanation 1:

`2Na+Cl_2\rightarrow 2NaCl`

moles of NaCl =
`\frac{\text{given mass of NaCl}}{\text{molar mass of NaCl}}=(200 g)/(58.5 g/mol)=3.4188 mol`

According to reaction, 2 moles NaCl are obtained from 2 moles of Na, then 3.4188 moles of NaCl are obtained from:
`(1)/(1)* 3.4188` moles of Na.

Mass of Na = moles of sodium × Molar mass of sodium

`=23 g/mol* 3.4188 mol=78.632 g`

From the given option nearest answer to calculated answer is option (B) that is 78.65 g.

Explanation 2:

`2K+2H_2O\rightarrow 2KOH+H_2`

moles of potassium(K)=
`\frac{\text{given mass of K}}{\text{molar mass of K}}=(23.5 g)/(39 g/mol)=0.6025 mol`

According to reaction, 2 moles of K gives 2 moles of KOH , then 0.6025 moles of K will give :
`(1)/(1)* 0.6025 moles` of KOH

Mass of KOH = moles of KOH × Molar mass of KOH

`0.6025 mol* 56 g/mol=33.74 g\approx 33.7 g`

From the given option nearest answer to calculated answer is option (A) that 33.7 g.