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3 years ago

Calculate the volume in ml of 0.200 M na2co3 needed to produce 2.00 g of caco3 there is an excess of cacl2

Chemistry

1 answer:

UkoKoshka [18]3 years ago

8 0

Answer:

$V=100mL$

Explanation:

Hello.

In this case, since the chemical reaction is:

$Na_2CO_3+CaCl_2\rightarrow CaCO_3+2NaCl$

We next compute the moles of sodium carbonate from the 2.00 grams of calcium carbonate via their 1:1 mole ratio in the chemical reaction:

$n_{Na_2CO_3}=2.00gCaCO_3*\frac{1molCaCO_3}{100.09gCaCO_3}*\frac{1molNa_2CO_3}{1molCaCO_3} \\\\n_{Na_2CO_3}=0.0200molNa_2CO_3$

Thus, by knowing the molarity, we compute the volume:

$M=\frac{n}{V}\\ \\V=\frac{n}{M}=\frac{0.0200mol}{0.200mol/L}\\ \\V=0.100L*\frac{1000mL}{1L}\\ \\V=100mL$

Best regards.

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Complete question:

Imagine two solutions with the same concentration and the same boiling point, but one has benzene as the solvent and the other has carbon tetrachloride as the solvent. Determine that molal concentration, m (or b), and boiling point, Tb.

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Tb - 76.8 = m*5.03*1

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