Question

question 1 (2 points)
answer the questions about the balanced reaction 3a + 2b → 1c + 100. cal
the molar masses are as follows: a = 10. g, b = 20. g, c = 70. g
how many grams of a are required to make 1 mole of c?
1 mol a
(1 mol c)(-------------) = 2
3 mol c
4 g a
(5 mol a)(-------------) = 6 g a
7 mol a
is the relationship between moles of c and grams of a a direct proportion? 8
a. 0 b. 1 c. 2 d. 3 e. 4 f. 5 g. 6 h. 7 i. 10
j. 20 k. 30 l. 70 m. 100 n. 0.143 o. 14.3 p. 0.286
q. 0.429 r. 4.29 s. 5.71 t. yes u. no

Explanation:

Step1: Determine moles of A from moles of C

From the balanced reaction \(3A + 2B
ightarrow1C+100\) cal, the mole ratio of \(A\) to \(C\) is \(3:1\). So, if we have \(1\) mol of \(C\), the moles of \(A\) required is \(3\) (from the coefficient ratio in the balanced equation).

Step2: Calculate grams of A

The molar mass of \(A\) is \(10\) g/mol. Using the formula \(m = n\times M\) (where \(m\) is mass, \(n\) is moles, and \(M\) is molar mass), for \(n = 3\) mol of \(A\) and \(M=10\) g/mol, we have \(m=(3\space mol)\times(10\space g/mol)\).

Step3: Check direct - proportion relationship

The formula for the mass of \(A\) (\(m_A\)) in terms of moles of \(C\) (\(n_C\)) is \(m_A=(n_C)\times(3\space mol\space A/1\space mol\space C)\times(10\space g\space A/1\space mol\space A)=30n_C\). Since \(m_A = k\times n_C\) (where \(k = 30\) is a constant), the relationship is a direct proportion.

Answer:

1. D. 3
2. \(3\)
3. B. 1
4. I. 10
5. D. 3
6. K. 30
7. B. 1
8. T. yes