QQuestionChemistry
QuestionChemistry
7. The molar mass of C is 22.99 g/mol and for C is it 12.01 g/mol. You just calculated that there would be 0.085 mols of Na and 0.085 mols of C. Calculate total mass of the sodium and carbon produced in this case. Hide Hint Hint: For this reaction there are two solids, Na and C. You'll need to calculate the mass of each of these and add them to find the total mass of the remaining solids. 2. Assume the reaction is: 2 NaHCO, (s) → Na,Co, (s) + CO^2(g) + H^2O(g) What mass of NaHCO, (s) must have been present at the beginning of the reaction? Hide Hint Hint: Your goal is to do the steps you'd did in the previous section, but in reverse. You know the mass of the Na,CO, product: 3.78g The molar mass of Na,Co, is 105.99 g/mol The molar mass of NaHCO, is 84.01 g/mol How can you use this information to find the mass of the initial amount of NaHCO? 0.042496559 g Submit Answer x unlimited submissions remaining Score: Comments: 0 / 1 Your goal is to do the steps you'd did in the previous section, but in reverse. You know the mass of the Na^2CO, product. How can you use this to find the mass of the initial amount of NaHCO,?
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Answer
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Step 1Let's solve each part step-by-step.
--- ### Part 1: Calculate total mass of sodium (Na) and carbon (C) produced **Given:** - Moles of Na = 0.085 mol - Moles of C = 0.085 mol - Molar mass of Na = 22.99 g/mol - Molar mass of C = 12.01 g/mol --- **Step 1:** Calculate the mass of sodium (Na). The formula to calculate mass from moles is: \text{mass} = \text{moles} \times \text{molar mass} So, --- **Step 2:** Calculate the mass of carbon (C). Similarly, --- **Step 3:** Calculate the total mass of sodium and carbon. Add the masses: m_{\text{total}} = m_{\text{Na}} + m_{\text{C}} = 1.95415 \, \text{g} + 1.02085 \, \text{g} = 2.975 \, \text{g} --- ### Part 2: Calculate the mass of NaHCO₃ initially present **Given:** --- Using the formula: So, --- From the balanced equation: 2 \, \text{mol NaHCO}_3 \rightarrow 1 \, \text{mol Na}_2\text{CO}_3 Therefore, n_{\text{NaHCO}_3} = 2 \times n_{\text{Na}_2\text{CO}_3} = 2 \times 0.03567 = 0.07134 \, \text{mol} --- Using the formula: m = n \times \text{molar mass} So, --- ###
Final Answer
- Total mass of sodium and carbon produced: \boxed{2.975 \, \text{g}} - Mass of \text{NaHCO}_3 initially present: \boxed{5.995 \, \text{g}}
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