Answer:
A. 1.28g H2CO3 are produced
B. 107% is percentage yield
Explanation:
Based on the reaction:
2NaHCO3 → Na2CO3 + H2CO3
<em>2 moles of NaHCO3 produce 1 mole of Na2CO3 and 1 mole of H2CO3</em>
A. To solve this question we need to find the moles of Na2CO3 = Moles of H2CO3. With their moles we can find the mass of H2CO3 as follows:
<em>Moles Na2CO3 -Molar mass: 105.99g/mol-</em>
2.19g Na2CO3 * (1mol/105.99g) = 0.02066 moles Na2CO3 = Moles H2CO3
<em>Mass H2CO3 -Molar mass: 62.03g/mol-</em>
0.02066 moles * (62.03g/mol) =
<h3>1.28g H2CO3 are produced</h3>
B. The percentage yield is the 100 times the ratio between actual yield (1.28g) and theoretical yield. The theoretical yield is obtained from the mass of NaHCO3 as follows:
<em>Moles NaHCO3 -Molar mass: 84.007g/mol-</em>
3.24g * (1mol / 84.007g) = 0.03857 moles NaHCO3
Assuming all these moles react producing H2CO3, the theoretical moles of H2CO3 are:
0.03857 moles NaHCO3 * (1mol H2CO3 / 2 mol NaHCO3) = 0.01928 moles NaHCO3
<em>Theoretical mass:</em>
0.01928 moles NaHCO3 * (62.03g/mol) = 1.20g of H2CO3 is theoretical mass.
<em>Percentage yield:</em>
1.28g H2CO3 / 1.20g H2CO3 =
<h3>107% is percentage yield</h3>