Na2CO3 + 2Cl- ⇒ 2NaCl + CO3^-2
<span>
1 mole of Na2CO3 = 106 g </span>
<span>2 moles of NaCl = 2 x 58.4
= 116.8 g
</span>Na2CO3 would increase by 116.8 / 106 = 1.10 to form 2NaCl.
<span>0.4862 g x 1.10 = 0.515 grams of NaCl.
</span>
K2CO3 + 2Cl- ⇒ 2KCl + CO3^-2
<span>1 mole of K2CO3 = 138.2 g </span>
<span>2 moles of KCl = 149.1 </span>
<span>
K2CO3 would increase by </span>149.1 /138.2 = 1.079 <span>to form 2KCl
</span>
<span> 0.4862 x 1.079 = 0.5246 g</span>
Answer:
Limiting - Na Excess - Cl2
Explanation:
the first is hydroxide. I believe d is the second one but I'm not 100% positive so you may want to get a second opinion
A. BeCl2 sp2
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