SOLUTIONS AND PURE SUBSTANCES TO2/S04 If 150 mL of 5.40 M aqueous HCI and 160 mL of 0.938 M aqueous Na2CO3 are reacted stoichiometrically according to the balanced equation, how many grams of liquid H20 are produced? Molar Mass (g/mol) 36.461 HCI NazCO3(aq) + 2HCI(aq) - CO2(g) + 2NACI(aq) + H2o(1) Na2CO3 105.99 Н2о 18.015 Density (g/mL) H20 0.9982 Molar Volume (L) 22.4 at STP Gas Constant (L'atm'mol"1-K*1) 0.0821 the tolerance is +/-2%

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SOLUTIONS AND PURE SUBSTANCES TO2/S04
If 150 mL of 5.40 M aqueous HCI and 160 mL of 0.938 M aqueous Na2CO3 are reacted stoichiometrically according to the balanced equation, how many grams of
liquid H20 are produced?
Molar Mass
(g/mol)
36.461
HCI
NazCO3(aq) + 2HCI(aq) - CO2(g) + 2NACI(aq) + H2o(1)
Na2CO3 105.99
Н2о
18.015
Density (g/mL)
H20 0.9982
Molar Volume (L)
22.4 at STP
Gas Constant
(L'atm'mol"1-K*1)
0.0821
the tolerance is +/-2%
Transcribed Image Text:SOLUTIONS AND PURE SUBSTANCES TO2/S04 If 150 mL of 5.40 M aqueous HCI and 160 mL of 0.938 M aqueous Na2CO3 are reacted stoichiometrically according to the balanced equation, how many grams of liquid H20 are produced? Molar Mass (g/mol) 36.461 HCI NazCO3(aq) + 2HCI(aq) - CO2(g) + 2NACI(aq) + H2o(1) Na2CO3 105.99 Н2о 18.015 Density (g/mL) H20 0.9982 Molar Volume (L) 22.4 at STP Gas Constant (L'atm'mol"1-K*1) 0.0821 the tolerance is +/-2%
Expert Solution
Step 1

Determine number of moles of each reactant:

1 mL = 0.001 L

Chemistry homework question answer, step 1, image 1

From the balanced equation, it is clear that 1 mol Na2CO3 reacts with 2 mol HCl.

Hence, 0.810 mol HCl has to react with 0.405 mol Na2CO3 but has only 0.150 mol Na2CO3. Therefore, Na2CO3 is the limiting reagent (excess HCl is present).

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