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Question
- in this experiment, you will be mixing aqueous solutions of sodium carbonate and calcium chloride to produce solid calcium carbonate.
na₂co₃(aq) + cacl₂(aq) → 2nacl(aq) + caco₃(s)
order the steps required to predict the volume (in ml) of 0.200 m calcium chloride needed to produce 2.00 g of calcium carbonate. there is an excess of sodium carbonate.
step 1
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compute the volume of calcium chloride solution required
compare moles of calcium carbonate to moles of calcium chloride based on balanced equation to calculate moles of calcium...
convert mass of calcium carbonate to moles of calcium carbonate
convert the volume of calcium chloride solution required from liters to milliliters
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step 4
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Step1: Convert mass of calcium carbonate to moles of calcium carbonate
$$n_{CaCO_3}=\frac{m_{CaCO_3}}{M_{CaCO_3}}$$
where \(m_{CaCO_3} = 2.00\space g\) and \(M_{CaCO_3}=(40.08 + 12.01+3\times16.00)\space g/mol = 100.09\space g/mol\)
Step2: Compare moles of calcium carbonate to moles of calcium chloride based on balanced equation to calculate moles of calcium chloride
From the balanced equation \(Na_2CO_3(aq)+CaCl_2(aq)\to2NaCl(aq)+CaCO_3(s)\), the mole ratio \(n_{CaCl_2}:n_{CaCO_3}=1:1\), so \(n_{CaCl_2}=n_{CaCO_3}\)
Step3: Compute the volume of calcium chloride solution required
Using the formula \(V=\frac{n}{C}\), where \(C = 0.200\space M\) (molarity of \(CaCl_2\) solution) and \(n=n_{CaCl_2}\)
Step4: Convert the volume of calcium chloride solution required from liters to milliliters
Since \(1\space L = 1000\space mL\), \(V_{mL}=V_{L}\times1000\)
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- Convert mass of calcium carbonate to moles of calcium carbonate
- Compare moles of calcium carbonate to moles of calcium chloride based on balanced equation to calculate moles of calcium chloride
- Compute the volume of calcium chloride solution required
- Convert the volume of calcium chloride solution required from liters to milliliters