Calculate the molarity of 150.0 mL of sodium hydroxide solution needed to react with 65.7 mL of a 0.128 M solution of phosphoric acid. OA. 0.877 M Ов.0.168 м OC. 0.0974 M OD.0.0420 M OE. 0.0187 M OF. 0.0390 M OG. 0.0561 M OH.0.292 M

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Chapter1: Chemical Foundations
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**Question 1**

Calculate the molarity of 150.0 mL of sodium hydroxide solution needed to react with 65.7 mL of a 0.128 M solution of phosphoric acid.

- A. 0.877 M
- B. 0.128 M
- C. 0.0974 M
- D. 0.0420 M
- E. 0.0187 M
- F. 0.0390 M
- G. 0.0564 M
- H. 0.292 M

**Question 2**

What volume of 0.128 M solution of hydrochloric acid is needed to react with 200.0 mL of 0.128 M calcium hydroxide?

- A. 0.0000410 mL
- B. 100 mL
- C. 0.00000419 mL
- D. 24400 mL
- E. 400 mL
- F. 0.00000819 mL
- G. 200 mL
- H. 6100 mL

**Question 3**

Given a density of 1.17 g/mL, what volume of a 50.5% solution of sulfuric acid is needed to react 350.0 mL of 0.138 M strontium hydroxide?

- A. 9.56 mL
- B. 13100 mL
- C. 5.15 mL
- D. 60.2 mL
- E. 3.8 mL
- F. 8.02 mL
- G. 13500 mL
- H. 4.40 mL

**Question 4**

Given 200.0 mL of a 0.139 M solution of aluminum chloride, according to the following equation, how many mL of a 0.358 M solution of sodium carbonate are needed to react with the aluminum chloride?

\[ 3 \text{ Na}_2\text{CO}_3(\text{aq}) + 2 \text{ AlCl}_3(\text{aq}) \rightarrow \text{ Al}_2(\text{CO}_3)_3(\text{s}) + 6 \text{ NaCl}(\text{aq}) \]

- A. 77.
Transcribed Image Text:**Question 1** Calculate the molarity of 150.0 mL of sodium hydroxide solution needed to react with 65.7 mL of a 0.128 M solution of phosphoric acid. - A. 0.877 M - B. 0.128 M - C. 0.0974 M - D. 0.0420 M - E. 0.0187 M - F. 0.0390 M - G. 0.0564 M - H. 0.292 M **Question 2** What volume of 0.128 M solution of hydrochloric acid is needed to react with 200.0 mL of 0.128 M calcium hydroxide? - A. 0.0000410 mL - B. 100 mL - C. 0.00000419 mL - D. 24400 mL - E. 400 mL - F. 0.00000819 mL - G. 200 mL - H. 6100 mL **Question 3** Given a density of 1.17 g/mL, what volume of a 50.5% solution of sulfuric acid is needed to react 350.0 mL of 0.138 M strontium hydroxide? - A. 9.56 mL - B. 13100 mL - C. 5.15 mL - D. 60.2 mL - E. 3.8 mL - F. 8.02 mL - G. 13500 mL - H. 4.40 mL **Question 4** Given 200.0 mL of a 0.139 M solution of aluminum chloride, according to the following equation, how many mL of a 0.358 M solution of sodium carbonate are needed to react with the aluminum chloride? \[ 3 \text{ Na}_2\text{CO}_3(\text{aq}) + 2 \text{ AlCl}_3(\text{aq}) \rightarrow \text{ Al}_2(\text{CO}_3)_3(\text{s}) + 6 \text{ NaCl}(\text{aq}) \] - A. 77.
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