1)A chemist wants to make 6.0 L of a 0.390 M CaCl2solution. What mass of CaCl2 (

Question

1)A chemist wants to make 6.0 L of a 0.390 M CaCl2solution.

What mass of CaCl2 (in g) should the chemist use?

2)To what volume should you dilute 30 mL of a 11.0 M H2SO4 solution to obtain a 0.170 M H2SO4 solution? (in L)

3)Consider the precipitation reaction: Li2S(aq)+Co(NO3)2(aq)2LiNO3(aq)+CoS(s)

What volume of 0.160 M Li2S solution is required to completely react with 125 mL of 0.160 M Co(NO3)2?

4)What is the minimum amount of 6.8 M H2SO4 necessary to produce 27.3 g of H2 (g) according to the following reaction? 2Al(s)+3H2SO4(aq)Al2(SO4)3(aq)+3H2(g)

Express your answer using two significant figures.

5)Consider the precipitation reaction: 2Na3PO4(aq)+3CuCl2(aq)Cu3(PO4)2(s)+6NaCl(aq)

What volume of 0.180 M Na3PO4 solution is necessary to completely react with 87.1 mL of 0.122 M CuCl2?

Explanation / Answer

1. No. of moles solute required

Molarity = moles of solute / Liter of solution

0.390M/L = x / 6L

x = 0.390 M/L * 6L

x = 2.34 M

Required mass of CaCl2 = No of moles * Molecular weight

= 2.34 mol * 110.98 g/mol

= 259.6932 g

2. V1* N1 =V2 * N2

V1 = stock volume (30mL)

N1 = stock concentration (11 M)

V2 = required volume (here x)

N2 = required concentration(0.17 M)

30mL * 11 M = x * 0.17 M

x = 1941.17 mL

x = 1.941 L

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